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Q15: The trajectory of a projectile in a vertical plane is $y=ax-bx^{2}$, where $a, b$ are constants, and $x$ and $y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is. and the angle of projection from the horizontal is
|
Theta=\tan ^{-1} a$$
|
$y=ax-bx^{2}$
Comparing the given equation with
$$y=(\tan \theta)x-\left(\frac{g}{2 u^{2} \cos ^{2} \theta}\right) x^{2}$$
Then $a=\tan \theta$ and
$$b=\frac{g}{2 u^{2} \cos ^{2} \theta}$$
Maximum height $=a^{2} / 4b$ and angle of projection,
$$\tan \Theta=a$$
$$\Theta=\tan ^{-1} a$$
|
Kinematics 2D
|
|
Q1: Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section $5 \mathrm{~mm}^{2}$ is $\mathrm{v}$. If the electron density of copper is $9 \times 10^{28} / \mathrm{m}^{3}$ the value of $v$ in $\mathrm{mm} / \mathrm{s$ is close to (Take charge of electron to be $=1.6 \times 10^{-19} \mathrm{C}$)
|
3
0.2
2
0.02
|
(d) 0.02
|
As $I=$ neAvd $=$ neAv
$v=1 / n e A=1.5 /\left(9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}\right)$
$v=0.02 \times 10^{-3} \mathrm{~m} / \mathrm{s}=0.02 \mathrm{~mm} / \mathrm{s$
|
Current Electricity
|
Q2: Two equal resistances when connected in series to a battery, consume electric power of $60 \mathrm{~W}$. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be
|
$240 \mathrm{~W}$
$120 \mathrm{~W}$
$60 \mathrm{~W}$
$30 \mathrm{~W}$
|
(a) $240 \mathrm{~W}$
|
The power consumed when two resistance are in series combination is $\mathrm{V}^{2} / 2 \mathrm{R}=60 \mathrm{~W} \Rightarrow \mathrm{V}^{2} / \mathrm{R}=120 \mathrm{~W}$
When the two resistance are connected in parallel combination, power consumed is $2 \mathrm{~V}^{2} / \mathrm{R}=120(2)=240 \mathrm{~W}$
|
Current Electricity
|
Q3: A current of $2 \mathrm{~mA}$ was passed through an unknown resistor which dissipated a power of $4.4 \mathrm{~W}$. Dissipated power when an ideal power supply of $11 \mathrm{~V}$ is connected across it is
|
$11 \times 10^{-4} \mathrm{~W}$
$11 \times 10^{-5} \mathrm{~W}$
$11 \times 10^{5} \mathrm{~W}$
$11 \times 10^{-3} \mathrm{~W}$
|
(b) $11 \times 10^{-5} \mathrm{~W}$
|
Case (1)
As $I^{2} R=P$
$R=P / I^{2}$
$R=(4.4) /\left(2 \times 10^{.3}\right)^{2}=1.1 \times 10^{6} \Omega$
Case (2)
$P=V^{2} / R=(1.1) 2 /\left(1.1 \times 10^{6}\right)=11 \times 10^{-5} \mathrm{~W}$
|
Current Electricity
|
Q4: An ideal battery of $4 \mathrm{~V}$ and resistance $\mathrm{R}$ are connected in series in the primary circuit of a potentiometer of length $1 \mathrm{~m}$ and resistance 5 . The value of $R$, to give a potential difference of $5 \mathrm{mV}$ across $10 \mathrm{~cm}$ of potentiometer wire is
|
490
495
395
480
|
(c) 395
|
Let I be the current in the circuit. $4=(5+R)$ \qquad
According to given condition,
$5 \times 10^{-3}=(10 / 100)(5)(I)$
$\mathrm{I}=10^{-2} \mathrm{~A}$
Using (1) and (2), $5+R=400 R=395$
|
Current Electricity
|
Q5: A cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when
|
$R=0.001 r$
$R=r$
$R=2 r$
$R=1000 r$
|
(b) $\mathrm{R}=\mathrm{r}$
|
The power delivered to resistance is $I^{2} R$
i.e., $P=\left[\varepsilon^{2} /(R+r)^{2}\right] R$
For the maximum power, $\mathrm{dP} / \mathrm{dR}=0$
$\Rightarrow-2 R+(R+r)=0$ or $R=r$
|
Current Electricity
|
Q6: A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^{\circ}$ at the centre, the equivalent resistance between these two points will be
|
$(7 / 2) \Omega$
$(5 / 2) \Omega$
$(12 / 5) \Omega$
$(5 / 3) \Omega$
|
(d) (5/3) $\Omega$
|
$R=3 \Omega=\rho(1 / A)=\rho\left(I^{2} / V\right)$
$R^{\prime}=\rho\left(l^{\prime} / V\right) \Rightarrow R^{\prime}=(21)^{2} / /^{2} \times 3 \Rightarrow R^{\prime}=12 \Omega$
Equivalent resistance, $\mathrm{R}_{\text {eq }}=(10 \times 2) /(10+2)=(5 / 3) \Omega$
|
Current Electricity
|
Q7: On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10 \mathrm{~cm}$. The resistance of the combination is $1 \mathrm{k} \Omega$. How much was the resistance on the left slot before the interchange?
|
990
505
550
910
|
(c) 550
|
Let $\mathrm{R}_{1}$ (left slot) and $\mathrm{R}_{2}$ (right slot) be two resistances in two slots of a meter bridge. Initially I be the balancing length
Then, $\mathrm{R}_{1} / \mathrm{R}_{2}=\mathrm{I} /(100-\mathrm{I})$
$\mathrm{R}_{1}+\mathrm{R}_{2}=1000 \Omega$
On interchanging the resistances, balancing length becomes (I -10 ), so
$\left(R_{2} / R_{1}\right)=(I-10) /(110-I)$
Using (1)
$(100-I) / I=(I-10) /(110-I)$
$11000+\left.\right|^{2}-210|=|^{2}-10 \mid$
$200 \mathrm{I}=11000, \mathrm{I}=55 \mathrm{~cm}$
From equa (1) $R_{1} / R_{2}=55 / 45$
Using (2)
$\mathrm{R}_{1}=(55 / 45)\left(1000-\mathrm{R}_{1}\right)$
$\mathrm{R}_{1}+(55 / 45) \mathrm{R}_{1}=(1000) \times(55 / 45)$
$100 \mathrm{R}_{1}=1000 \times 55$
$\mathrm{R}_{1}=550 \Omega$
|
Current Electricity
|
Q8: A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be
|
Increased 8 times
Unchanged
Doubled
Halved
|
(a) Increased 8 times
|
Rate of heat developed, $\mathrm{P}=\mathrm{V}^{2} / \mathrm{R}$
For given $V, P \propto 1 / R=A / \rho l=\pi r^{2} / \rho l$
Now, $\mathrm{P}_{1} / \mathrm{P}_{2}=\left(\mathrm{r}_{1}^{2} / \mathrm{r}_{2}^{2}\right)\left(\mathrm{I}_{2} / I_{1}\right)$
As per question, $l_{2}=I_{1} / 2$ and $r_{2}=2 r_{1}$
$P_{1} / P_{2}=(1 / 4) \times(1 / 2)=1 / 8$
$\mathrm{P}_{2}=8 \mathrm{P}_{1}$
|
Current Electricity
|
Q9: A heating element has a resistance of $100 \Omega$ at room temperature. When it is connected to a supply of $220 \mathrm{~V}$, a steady current of $2 \mathrm{~A}$ passes in it and the temperature is $500^{\circ} \mathrm{C}$ more than room temperature. What is the temperature coefficient of resistance of the heating element?
|
$1 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
$2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
$0.5 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
$5 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
|
(b) $2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
|
Resistance after temperature increases by $500^{\circ} \mathrm{C}$,
$\mathrm{R}_{\mathrm{T}}=$ Voltage applied/Current $=220 / 2=110$
Also, $\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$
$110=100(1+(\alpha \times 500))$
$\alpha=10 /(100 \times 500)=2 \times 10^{-4} \mathrm{C}^{-1}$
|
Current Electricity
|
Q10: A uniform wire of length I and radius $r$ has a resistance of $100 \Omega$. It is recast into a wire of radius $\mathrm{r} / 2$.The resistance of new wire will be
|
$400 \Omega$
$100 \Omega$
$200 \Omega$
$1600 \Omega$
|
(d) $1600 \Omega$
|
Resistance of a wire of length I and radius $r$ is given by
$R=\rho I / A=(\rho I / A) x(A / A)$
$\mathrm{R}=\left(\rho \mathrm{I} / \mathrm{A}^{2}\right)=\left(\rho \mathrm{V} / \pi^{2} \mathrm{r}^{4}\right)(\because \mathrm{V}=\mathrm{Al})$
i.e., $R \propto 1 / r^4$
$R_{1} / R_{2}=\left(r_{2} / r_{1}\right)^{4}$
Here, $R_{1}=100 \Omega, r_{1}=r, r_{2}=r / 2$
$R_{2}=R_{1}\left(r_{1} / r_{2}\right)^{4}=16 R_{1}=1600 \Omega$
|
Current Electricity
|
Q11: A 2 W carbon resistor is colour coded with green, black, red and brown, respectively. The maximum current which can be passed through this resistor is
|
$20 \mathrm{~mA}$
$0.4 \mathrm{~mA}$
$100 \mathrm{~mA}$
$63 \mathrm{~mA}$
|
(a) $20 \mathrm{~mA}$
|
The resistance of the resistor is $50 \times 10^{2} \Omega$. So, the maximum current that can be passed through it is
$$
\sqrt{\frac{P}{R}}=\sqrt{\frac{2}{50 \times 10^{2}}} A=20 \mathrm{~mA}$$
|
Current Electricity
|
Q12: In a large building, there are 15 bulbs of $40 \mathrm{~W}, 5$ bulbs of $100 \mathrm{~W}, 5$ fans of $80 \mathrm{~W}$ and 1 heater of 1 kW. The voltage of the electric mains is $220 \mathrm{~V}$. The minimum capacity of the main fuse of the building will be
|
$14 \mathrm{~A}$
$8 \mathrm{~A}$
$10 \mathrm{~A}$
$12 \mathrm{~A}$
|
(d) $12 \mathrm{~A}$
|
Power of 15 bulbs of $40 \mathrm{~W}=15 \times 40=600 \mathrm{~W}$
Power of 5 bulbs of $100 \mathrm{~W}=5 \times 100=500 \mathrm{~W}$
Power of 5 fans of $80 \mathrm{~W}=5 \times 80=400 \mathrm{~W}$
Power of 1 heater of $1 \mathrm{~kW}=1000$
Total power, $P=600+500+400+1000=2500 \mathrm{~W}$
When these combinations of bulbs, fans and heater are connected to $220 \mathrm{~V}$ mains, current in the main fuse of the building is given by
$\mathrm{I}=\mathrm{P} / \mathrm{N}=2500 / 220=11.36 \mathrm{~A} \approx 12 \mathrm{~A}$
|
Current Electricity
|
Q13: If a wire is stretched to make it $0.1 \%$ longer, its resistance will
|
increase by $0.05 \%$
increase by $0.2 \%$
decrease by $0.2 \%$
decrease by $0.05 \%$
|
(b) increase by $0.2 \%$
|
Resistance of wire $\mathrm{R}=\rho / / \mathrm{A}$
On stretching, volume $(\mathrm{V})$ remains constant.
So $\mathrm{V}=\mathrm{Al}$ or $\mathrm{A}=\mathrm{V} / \mathrm{I}$
Therefore, $R=\rho l^{2} / V$ (Using (1))
Taking logarithm on both sides and differentiating we get,
$\Delta R / R=2 \Delta I / I$ (Since $V$ and $\rho$ are constants)
$(\Delta R / R) \%=(2 \Delta I / l) \%$
Hence, when wire is stretched by $0.1 \%$ its resistance will increase by $0.2 \%$
|
Current Electricity
|
Q14: A thermocouple is made from two metals, antimony and bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will
|
flow from antimony to bismuth at the cold junction
flow from antimony to bismuth at the hot junction
flow from bismuth to antimony at the cold junction
not flow through the thermocouple.
|
(a) flow from antimony to bismuth at the cold junction
|
Antimony-bismuth couple is $A B C$ couple. It means that current flows from $A$ to $B$ at a cold junction
|
Current Electricity
|
Q15: The resistance of a wire is 5 ohm at $50^{\circ} \mathrm{C}$ and 6 ohm at $100^{\circ} \mathrm{C}$. The resistance of the wire at $0^{\circ} \mathrm{C}$ will be
|
3 ohm
2 ohm
1 ohm
4 ohm
|
(d) 4 ohm
|
$R_{t}=R_{0}(1+\alpha t)$
$R_{t}$ is the resistance of wire at $t^{0} \mathrm{C}$
$R_{0}$ is the resistance of wire at $0^{\circ} \mathrm{C}$
$\alpha$ is the temperature coefficient of resistance
$R_{50}=R_{0}[1+\alpha(50)]$
And $R_{100}=R_{0}[1+\alpha(100)]$
Or $\mathrm{R}_{50}-\mathrm{R}_{0}=\mathrm{R}_{0} \alpha(50)$
$R_{100}-R_{0}=R_{0} \alpha(100)$
Dividing (1) by (2)
$\left(5-\mathrm{R}_{0}\right) /\left(\left(6-\mathrm{R}_{0}\right)=1 / 2\right.$
$\mathrm{R}_{0}=4$ ohm
|
Current Electricity
|
Q1: A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity 100 m / s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g=10 m / s)
|
(a) 10 m
(b) 30 m
(c) 20 m
(d) 40 m
|
(d) 40 m
|
Suppose both collide at the point P after time t. Time taken for the particles to collide, t=d / v_rel = 100 / 100 = 1 s
Speed of wood just before collision = g t = 10 m / s
Speed of bullet just before collision v - g t = 100 - 10 = 90 m / s
Before 0.03 kg ↓ 10 m / s
0.02 kg ↑ 90 m / s
After 个 v 0.05 kg
Now, the conservation of linear momentum just before and after the collision - (0.03)(10) + (0.02)(90) = (0.05) v ⇒ v = 30 m / s
The maximum height reached by the body a = v² / 2 g = (30)² / 2(10) = 45 m
(100 - h) = 1 / 2 g t² = 1 / 2 × 10 × 1 ⇒ h = 95 m
Height above tower = 40 m
|
Kinematics 1D
|
Q2: A passenger train of length 60 m travels at a speed of 80 km / hr. Another freight train of length 120 m travels at a speed of 30 km / hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction, is
|
(a) 25 / 11
(b) 3 / 2
(c) 5 / 2
(d) 11 / 5
|
(d) 11 / 5
|
The total distance to be travelled by train is 60 + 120 = 180 m.
When the trains are moving in the same direction, the relative velocity is v₁ - v₂ = 80 - 30 = 50 km hr⁻¹. So time taken to cross each other, t₁ = 180 / (50 × 10³ / 3600) = [(18 × 18) / 25] s
When the trains are moving in opposite direction, relative velocity is |v₁ - (- v₂)| = 80 + 30 = 110 km hr⁻¹
So time taken to cross each other t₂ = 180 / (110 × 10³ / 3600) = [(18 × 36) / 110] s
t₁ / t₂ = [(18 × 18) / 25] / [(18 × 36) / 110] = 11 / 5
|
Kinematics 1D
|
Q3: A particle has an initial velocity 3\hat{i}+4\hat{j} and an acceleration of 0.4\hat{i}+0.3\hat{j}. Its speed after 10 s is
|
(a) 7 units
(b) 8.5 units
(c) 10 units
(d) 7\sqrt{2} units
|
(d) 7\sqrt{2} units
|
v = u + a t
u = 3 i + 4 j + (0.4 i + 0.3 j) × 10 = 3 i + 4 j + 4 i + 3 i
u = 7 i + 7 j
|u| = \sqrt{7² + 7²}
|
Kinematics 1D
|
Q4: An automobile, travelling at 40 km / h, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km / h, the minimum stopping distance, in metres, is (assume no skidding)
|
(a) 100 m
(b) 75 m
(c) 160 m
(d) 150 m
|
(c) 160 m
|
Using v² = u² - 2 a s
0 = u² - 2 a s
S = u² / 2 a
S₁ / S₂ = u₁² / u₂²
S₂ = (u₁² / u₂²) S₁ = (2)² (40) = 160 m
|
Kinematics 1D
|
Q7. A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m / s². He reaches the ground with a speed of 3 m / s. At what height, did he bailout?
|
(a) 293 m
(b) 111 m
(c) 91 m
(d) 182 m
|
(a) 293 m
|
Initially, the parachutist falls under gravity u² = 2 a h = 2 × 9.8 × 50 = 980 m² s⁻²
He reaches the ground with speed = 3 m / s,
a = -2 m s⁻² ⇒ (3)² = u² - 2 × 2 × h₁
9 = 980 - 4 h₁
h₁ = 971 / 4
h₁ = 242.75 m
Total height = 50 + 242.75 = 292.75 = 293 m.
|
Kinematics 1D
|
Q8: A car, starting from rest, accelerates at the rate f through a distance s, then continues at a constant speed for time t and then decelerates at the rate f / 2 to come to rest. If the total distance traversed in 15 s, then
|
(a) s = 1 / 2 ft²
(b) s = (1 / 4) ft²
(c) s = ft
(d) s = (1 / 72) ft²
|
(d) s = (1 / 72) ft²
|
For the first part of the journey, s = s₁,
s₁ = 1 / 2 ft₁²
v = f t₁
For second part of journey, s₂ = v t or s₂ = f t₁ t
For the third part of the journey, s₃ = 1 / 2(f / 2)(2 t₁)² = 1 / 2 × ft₁²
s₃ = 2 s₁ = 2 s
s₁ + s₂ + s₃ = 15 s
Ors s + ft₁ t + 2 s = 15 s
ft₁ t = 12 s
From (1) and (5) we get (s / 12 s) = ft₁² / (2 × ft₁ t)
Or t₁ = t / 6
Or s = 1 / 2 ft₁²
s = 1 / 2 f(t / 6)²
s = ft² / 72
|
Kinematics 1D
|
Q10: Which of the following statements is false for a particle moving in a circle with a constant angular speed?
|
(a) The velocity vector is tangent to the circle
(b) The acceleration vector is tangent to the circle
(c) The acceleration vector points to the centre of the circle
(d) The velocity and acceleration vectors are perpendicular to each other
|
(b) The acceleration vector acts along the radius of the circle. The given statement is false.
|
Kinematics 1D
|
|
Q11: From a building, two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then
|
(a) $v_{B}>v_{A}$
(b) $v_{A}=v_{B}$
(c) $v_{A}>v_{B}$
(d) their velocities depend on their masses
|
(b) $v_{A}=v_{B}$
|
Ball A projected upwards with velocity $u$, falls back with velocity $u$ downwards. It completes its journey to the ground under gravity.
$v_{A}^{2}=u^{2}+2 g h \ldots(1)$
Ball B starts with downwards velocity $u$ and reaches ground after travelling a vertical distance $h \quad v_{B} 2=$ $u^{2}+2 g h$
From (1) and (2)
$v_{A}=v_{B}$
|
Kinematics 1D
|
Q12: If a body loses half of its velocity on penetrating $3 \mathrm{~cm}$ in a wooden block, then how much will it penetrate more before coming to rest?
|
(a) $1 \mathrm{~cm}$
(b) $2 \mathrm{~cm}$
(c) $3 \mathrm{~cm}$
(d) $4 \mathrm{~cm}$
|
(a) $1 \mathrm{~cm}$
|
For first part of penetration, by equation of motion, $(u / 2)^{2}=u^{2}-2 a(3)$ $3 u^{2}=24 a \Rightarrow u^{2}=8 a$ For latter part of penetration, $0=(u / 2)^{2}-2 a x$ or $u^{2}=8 a x$.
From (1) and (2)
$8 \mathrm{ax}=8 \mathrm{a} \quad \mathrm{x}=1 \mathrm{~cm}$
|
Kinematics 1D
|
Q13: A car is standing $\mathbf{2 0 0} \mathrm{m}$ behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2 \mathrm{~m} / \mathrm{s}^{2}$ and the car has acceleration $4 \mathrm{~m} / \mathrm{s}^{2}$. The car will catch up with the bus after a time of
|
(a) $120 \mathrm{~s}$
(b) $15 \mathrm{~s}$
(c) [latex]2\textbackslash sqrt\{2\}[/latex] s
(d) $110 \mathrm{~s}$
|
(c) [latex]2\textbackslash sqrt\{2\}[/latex] s
|
Acceleration of car, $\mathrm{ac}_{\mathrm{c}}=4 \mathrm{~m} \mathrm{~s}^{-2}$
Acceleration of bus, $a_{B}=2 \mathrm{~m} \mathrm{~s}^{-2}$
Initial separation between the bus and car, $S_{C B}=200 \mathrm{~m}$
Acceleration of car with respect to bus, $a_{C B}=a_{c}-a_{B}=2 \mathrm{~m} \mathrm{~s}^{-2}$
Initial velocity $u_{C B}=0, t=$ ?
As, $s_{C B}=u_{C B} \times t+1 / 2 a_{C B} t^{2}$
$200=0 \times t+1 / 2 \times 2 \times t^{2}$
i.e., $\mathrm{t}^{2}=200$
$\mathrm{t}=[2\textbackslash sqrt{2}]$
|
Kinematics 1D
|
Q14: From a tower of height $H$, a particle is thrown vertically upwards with a speed $u$. The time taken by the particle, to hit the ground is $n$ times that taken by it to reach the highest point of its path. The relation between $\mathrm{H}, \mathrm{u}$ and $\mathrm{n}$ is
|
(a) $g H=(n-2) u^{2}$
(b) $2 g H=n 2 u^{2}$
(c) $g H=(n-2) 2 u^{2}$
(d) $2 g H=n u^{2}(n-2)$
|
(d) $2 g H=n u^{2}(n-2)$
|
Kinematics 1D
|
|
Q1: A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is
|
(a) 20 RT
(b) 12 RT
(c) 4 RT
(d) 15 RT
|
(d) 15 RT
|
$U=\left(f_{1} / 2\right) n_{1} R T+\left(f_{2} / 2\right) n_{2} R T$
$U=(5 / 2)(3 R T)+(3 / 2)(5 R T)$
$U=15 R T$
|
Kinetic Theory of Gases
|
Q2: An ideal gas is enclosed in a cylinder at a pressure of 2 atm and temperature, $300 \mathrm{~K}$. The mean time between two successive collisions is $6 \times 10^{-8} \mathrm{~s}$. If the pressure is doubled and the temperature is increased to $500 \mathrm{~K}$, the mean time between two successive collisions will be close to
|
(a) $4 \times 10^{-8} \mathrm{~s}$
(b) $3 \times 10^{-6} \mathrm{~s}$
(c) $0.5 \times 10^{-8} \mathrm{~S}$
(d) $2 \times 10^{-7} \mathrm{~s}$
|
(a) $4 \times 10^{-8} \mathrm{~s}$
|
Root mean square velocity
$$
\operatorname{Vrms}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 P}{\rho}}
$$
Here,
$R$ is the universal gas constant
$M$ is the molar mass
$\mathrm{P}$ is the pressure due to gas
$\rho$ is the density
Vrms $\propto \sqrt{T}$
$\mathrm{V}_{\mathrm{rms}} \propto$ mean free path/ time between successive collisions
And mean free path,
$$
Y=\frac{k T}{\sqrt{2 \pi \sigma^{2} P}}
$$
Vrms $\propto \mathrm{Y} / \mathrm{b}$
$$
\begin{aligned}
& \text { Vrms } \propto T / \sqrt{p} \times t---(1) \\
& \text { But Vrms } \propto \sqrt{T}----(2)
\end{aligned}
$$
$\sqrt{T} \propto T / \sqrt{p} \times t$
$\mathrm{t} \propto \sqrt{T} / p \quad \frac{t_{2}}{t_{1}}=\sqrt{\frac{T_{2}}{T_{1}} \times \frac{P_{1}}{P_{2}}} \frac{t_{2}}{t_{1}}$
$=\sqrt{\frac{500}{300} \times \frac{P_{1}}{2 P_{1}}}=\sqrt{\frac{5}{6}} t_{2}=\sqrt{\frac{5}{6}} t_{1}$
$\mathrm{t}_{2} \approx 4 \times 10^{-8} \mathrm{~s}$
|
Kinetic Theory of Gases
|
Q3: The specific heats, $C_{P}$ and $C_{V}$ of gas of diatomic molecules, $A$, is given (in units of $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then
|
(a) A has a vibrational mode but $\mathrm{B}$ has none.
(b) A has one vibrational mode and $B$ has two.
(c) $A$ is rigid but $B$ has a vibrational mode.
(d) Both $A$ and $B$ have a vibrational mode each.
|
(a) A has a vibrational mode but $B$ has none.
|
Here $\mathrm{Cp}$ and $\mathrm{Cv}$ of A are 29 and 22 and $\mathrm{Cp}$ and $\mathrm{Cv}$ of $\mathrm{B}$ are 30 and 21 .
$\nu=\mathrm{Cp} / \mathrm{Cv}=1+2 / \mathrm{f}$
For $\mathrm{A}, \mathrm{Cp} / \mathrm{Cv}=1+2 / \mathrm{f} \Rightarrow \mathrm{f}=6$
Molecules $A$ has 3 translational, 2 rotational and 1 vibrational degree of freedom
For $B, C p / C v=1+2 / f \Rightarrow f=5$
i.e., $B$ has 3 translational and 2 rotational degrees of freedom.
|
Kinetic Theory of Gases
|
Q4: Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? ( $R=8.3 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ )
|
(a) $21.6 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
(b) $19.7 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
(c) $15.7 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
(d) $17.4 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
|
(d) $17.4 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
|
$\mathrm{C}_{\mathrm{v} 1}$ of helium $=(3 / 2) \mathrm{R}$
$\mathrm{C}_{\mathrm{v} 2}$ of hydrogen $=(5 / 2) \mathrm{R}$
$$
\mathrm{C}_{\mathrm{v}} \text { of mixture }=\frac{2 \times \frac{3}{2} R+3 \times \frac{5}{2} R}{(2+3)}
$$
$C_{v}$ of mixture $=17.4 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
|
Kinetic Theory of Gases
|
Q5: The mass of a hydrogen molecule is $3.32 \times 10^{-27} / \mathrm{kg}$. If $10^{23}$ hydrogen molecules strike, per second, a fixed wall of area $2 \mathrm{~cm}^{2}$ at an angle of $45^{\circ}$ to the normal, and rebound elastically with a speed of $10^{3} \mathrm{~m} \mathrm{~s}-1$, then the pressure on the wall is nearly
|
(a) $2.35 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}$
(b) $4.70 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}$
(c) $2.35 \times 10^{2} \mathrm{~N} \mathrm{~m}^{-2}$
(d) $4.70 \times 10^{2} \mathrm{~N} \mathrm{~m}^{-2}$
|
(c) $2.35 \times 10^{2} \mathrm{~N} \mathrm{~m}^{-2}$
|
As $\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}$
Net force on the wall,
$\mathrm{F}=\mathrm{dp} / \mathrm{dt}=2 \mathrm{np} \mathrm{f}_{\mathrm{f}} \cos 45^{\circ}=2 \mathrm{nmv} \cos 45^{\circ}$
Here, $\mathrm{n}$ is the number of hydrogen molecules striking per second.
Pressure $=F / A=\left(2 n m v \cos 45^{\circ}\right) /$ Area
$$
=\frac{2 \times 10^{23} \times 3.32 \times 10^{-27} \times 10^{3} \times(1 / \sqrt{2})}{2 \times 10^{-4}}=2.35 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}
$$
|
Kinetic Theory of Gases
|
Q6: In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on $\mathrm{T}$ as $\mathrm{T}^{q}$. A good estimate for $q$ is
|
(a) 2
(b) 1
(c) $4 / 5$
(d) $4 / 7$
|
(b) 1
|
Pressure, $\mathrm{P}=1 / 3(\mathrm{mN} / \mathrm{V}) \mathrm{V}_{\text {rms }}^{2}$
$\mathrm{P}=(\mathrm{mN}) \mathrm{T} / \mathrm{V}$
If the gas mass and temperature are constant then
$\mathrm{P} \propto\left(\mathrm{V}_{\mathrm{rms}}\right)^{2} \propto \mathrm{T}$
So, force $\propto\left(\mathrm{V}_{\mathrm{rms}}\right)^{2} \propto \mathrm{T}$ I.e., Value of $q=1$
|
Kinetic Theory of Gases
|
Q7: One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
|
(a) $3 \times 10^{4} \mathrm{~J}$
(b) $5 \times 10^{4} \mathrm{~J}$
(c) $6 \times 10^{4} \mathrm{~J}$
(d) $7 \times 10^{4} \mathrm{~J}$
|
(b) $5 \times 10^{4} \mathrm{~J}$
|
The thermal energy or internal energy is $52 \mathrm{U} T$ for diatomic gases. (degree of freedom for diatomic gas $=5$ )
But PV $=$ RT
$V=$ mass $/$ density $=1 \mathrm{~kg} /\left(4 \mathrm{~kg} / \mathrm{m}^{3}\right)=(1 / 4) \mathrm{m}^{3}$
$P=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$
$U=(5 / 2) \times 8 \times 10^{4} \times 1 / 4=5 \times 10^{4} \mathrm{~J}$
|
Kinetic Theory of Gases
|
Q8: A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. The ratio $\mathrm{C}_{\mathrm{P}} / \mathrm{C}_{V}$ of the mixture is
|
(a) 1.4
(b) 1.54
(c) 1.59
(d) 1.62
|
(d) 1.62
|
For $16 \mathrm{~g}$ of helium, $\mathrm{n}_{1}=16 / 4=4$
For $16 \mathrm{~g}$ of oxygen, $\mathrm{n}_{2}=16 / 32=1 / 2$
For a mixture of gases
$\mathrm{C}_{\mathrm{v}}=\left(\mathrm{n}_{1} \mathrm{Cv}_{1}+\mathrm{n}_{2} \mathrm{Cv}_{2}\right) /\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)$ where $\mathrm{C}_{\mathrm{v}}=(\mathrm{f} / 2) \mathrm{R}$
$C_{p}=\left(n_{1} C_{1}+n_{2} C p_{2}\right) /\left(n_{1}+n_{2}\right)$ where $C_{p}=([f / 2]+1) R$
For helium, $f=3, n_{1}=4$
For oxygen, $f=5, n_{2}=1 / 2$
$$
\frac{C p}{C v}=\frac{\left(4 \times \frac{5}{2} R\right)+\left(\frac{1}{2} \times \frac{7}{2} R\right)}{\left(4 \times \frac{3}{2} R\right)+\left(\frac{1}{2} \times \frac{5}{2} R\right)}=\frac{47}{29}=1.62
$$
|
Kinetic Theory of Gases
|
Q9: Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will
|
(a) increase
(b) decrease
(c) remain the same
d) decrease for some, while the increase for others.
|
(c) remain the same
|
It is the relative velocities between molecules that are important. Root mean square velocities are different from lateral translation.
|
Kinetic Theory of Gases
|
Q10: At what temperature is the root mean square velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} \mathrm{C}$ ?
|
(a) $80 \mathrm{~K}$
(b) $-73 K$
(c) $3 \mathrm{~K}$
(d) $20 \mathrm{~K}$.
|
(d) $20 \mathrm{~K}$
|
$$
v_{r m s}=\sqrt{R T / M}\left(\mathrm{~V}_{\mathrm{rms}}\right) \mathrm{O}_{2}=\left(\mathrm{V}_{\mathrm{rms}}\right) \mathrm{H}_{2}
$$
$$\sqrt{\frac{273+47}{32}}=\sqrt{\frac{T}{2}}$$
$\mathrm{T}=20 \mathrm{~K}$
|
Kinetic Theory of Gases
|
Q11:The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to [Boltzmann constant, $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$, Avogadro Number, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{26} / \mathrm{kg}$, Radius of Earth: $6.4 \times 10^{6} \mathrm{~m}$, Gravitational acceleration on Earth $=10 \mathrm{~m} \mathrm{~s}^{-2}$ ]
|
(a) $10^{4} \mathrm{~K}$
(b) $800 \mathrm{~K}$
(c) $650 \mathrm{~K}$
(d) $3 \times 10^{5} \mathrm{~K}$
|
(a) $10^{4} \mathrm{~K}$
|
The escape speed of the molecule, $v_{e}=\sqrt{2 g R}$
Root mean square velocity, $v_{r m s}=\sqrt{\frac{\left.3\left(k_{B} N\right) T\right)}{m}}$
So, for $v_{e}=v_{r m s} \Rightarrow 2 g R=3\left(k_{B} \mathrm{~N}\right) \mathrm{T} / \mathrm{m}$
$\mathrm{T}=2 \mathrm{gRm} / 3 \mathrm{k}_{\mathrm{B}} \mathrm{N}=\left(2 \times 10 \times 6.4 \times 10^{6} \times 2 \times 10^{-3}\right) /\left(3 \times 1.38 \times 10^{-23} \times 6.02 \times 10^{23}\right)=10^{4} \mathrm{~K}$
|
Kinetic Theory of Gases
|
Q13: The value closest to the thermal velocity of a Helium atom at room temperature ( $300 \mathrm{~K}$ ) in $\mathrm{m} \mathrm{s}$ ${ }^{1}$ is $\left[k_{B}=1.4 \times 10^{-23} \mathrm{~J} / \mathrm{K} ; \mathrm{m}_{\mathrm{He}}=7 \times 10^{-27} \mathrm{~kg}\right]$
|
(a) $1.3 \times 10^{3}$
(b) $1.3 \times 10^{5}$
(c) $1.3 \times 10^{2}$
(d) $1.3 \times 10^{4}$
|
(a) $1.3 \times 10^{3}$
|
$$
\begin{aligned}
& \text { (3/2) } \mathrm{k}_{\mathrm{B}} \mathrm{T}=1 / 2 \mathrm{mv} \\
& v=\sqrt{\frac{3 k_{B} T}{m}}=\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}} \\
& =1.3 \times 10^{3} \mathrm{~ms}^{-1}
\end{aligned}
$$
|
Kinetic Theory of Gases
|
Q14: An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure ( $C_{P}$ ) and at constant volume ( $C_{V}$ ) is
|
(a) 6
(b) $7 / 2$
(c) $5 / 2$
(d) $7 / 5$
|
(d) $7 / 5$
|
An ideal gas has molecules with 5 degrees of freedom, then
$\mathrm{Cv}=(5 / 2) \mathrm{R}$ and $\mathrm{Cp}=(7 / 2) \mathrm{R}$
$\mathrm{Cp} / \mathrm{Cv}=(7 / 2) \mathrm{R} /(5 / 2) \mathrm{R}$
$\mathrm{Cp} / \mathrm{Cv}=7 / 5$
|
Kinetic Theory of Gases
|
Q15: $\mathbf{N}$ moles of a diatomic gas in a cylinder is at a temperature $\mathrm{T}$. Heat is supplied to the cylinder such that the temperature remains constant but $n$ moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas?
|
(a) 0
(b) $(5 / 2) n R T$
(c) $\left(\frac{1}{2}\right) \mathrm{nRT}$
(d) $(3 / 2) n R T$
|
(c) $(1 / 2)$ nRT
|
Initial kinetic energy of the system $K_{i}=(5 / 2)$ RTN
Final kinetic energy of the system $K_{f}=(5 / 2) R T(N-n)+(3 / 2) R T(2 n)$
$K_{f}-K_{i}=\Delta K=n R T(3-5 / 2)=(1 / 2) n R T$
|
Kinetic Theory of Gases
|
A paramagnetic material has $10^{28}$ atoms $/ \mathrm{m}^{3}$. It's magnetic susceptibility at temperature $350 \mathrm{~K}$ is $2.8 \times 10^{-4}$. Its susceptibility at $300 \mathrm{~K}$ is
|
(a) $3.267 \times 10^{-4}$, (b) $3.672 \times 10^{-4}$, (c) $2.672 \times 10^{-4}$, (d) $3.726 \times 10^{-4}$
|
(a) $3.267 \times 10^{-4}$
|
For a paramagnetic material, magnetic susceptibility $\chi \propto 1 / T$ $\chi_{2}=\chi_{1}\left(T_{1} / T_{2}\right)=2.8 \times 10^{-4} \times(350 / 300)=3.267 \times 10^{-4}$
|
Magnetism
|
A magnetic compass needle oscillates 30 times per minute at a place where the dip is $45^{\circ}$, and 40 times per minute where the dip is $30^{\circ}$. If $B_{1}$ and $B_{2}$ are respectively the total magnetic field due to the earth at the two places, then the ratio $B_{1} / B_{2}$ is best given by
|
(a) 0.7, (b) 3.6, (c) 1.8, (d) 2.2
|
(a) 0.7
|
Frequency of oscillation $=\sqrt{\text { magnetic field }}$ $\frac{30}{40}=\sqrt{\frac{B_{1} \cos 45^{\circ}}{B_{2} \cos 30^{\circ}}}$ (Horizontal component of magnetic field $=\mathrm{B} \cos \delta$ ) $\left(B_{1} / B_{2}\right)=0.7$
|
Magnetism
|
A magnetic needle of magnetic moment $6.7 \times 10^{-2} \mathrm{~A} \mathrm{~m}^{2}$ and moment of inertia $7.5 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^{2}$ is performing simple harmonic oscillations in a magnetic field of $0.01 \mathrm{~T}$. Time taken for 10 complete oscillations is
|
(a) $6.65 \mathrm{~s}$, (b) $8.89 \mathrm{~s}$, (c) $6.98 \mathrm{~s}$, (d) $8.76 \mathrm{~s}$
|
(a) $6.65 \mathrm{~s}$
|
Time period of magnetic needle oscillating simple harmonically is given by $T=2 \pi \sqrt{\frac{I}{M B}} T=2 \pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}$ $\mathrm{T}=(2 \pi / 10) \times 1.05 \mathrm{~s}$ For 10 oscillations, total time taken $\mathrm{T}^{\prime}=10 \mathrm{~T}=2 \pi \times 1.05=6.65 \mathrm{~s}$
|
Magnetism
|
A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\circ}$. One of the fields has a magnitude of $15 \mathrm{mT}$. The dipole attains stable equilibrium at an angle of $30^{\circ}$ with this field. The magnitude of the other field (in $\mathrm{mT}$ ) is close to
|
(a) 1, (b) 11, (c) 36, (d) 1060
|
(b) 11
|
The magnetic dipole attains stable equilibrium under the influence of these two fields making an angle $\theta_{1}=30^{\circ}$ with $B_{1}$ and $\theta_{2}=75^{\circ}-30^{\circ}=45^{\circ}$ with $B_{2}$. For stable equilibrium, net torque acting on dipole must be zero, I.e., $\tau_{1}+\tau_{2}=0$ $\tau_{1}=\tau_{2}$ $m B_{1} \sin \theta_{1}=m B_{2} \sin \theta_{2}$ $B_{2}=B_{1}\left(\sin \theta_{1} / \sin \theta_{2}\right)=15 \mathrm{mT} \times\left(\sin 30^{\circ} / \sin 45^{\circ}\right)$ $=15 \mathrm{mT} \times(1 / 2) \times \sqrt{2}=10.6 \mathrm{mT}=$
|
Magnetism
|
A short bar magnet is placed in the magnetic meridian of the earth with a north pole pointing north. Neutral points are found at a distance of $30 \mathrm{~cm}$ from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $\mathrm{Am}^{2}$ is close to (Given $\left(\mu_{0} / 4 \pi\right)=10^{-7}$ in SI units and $B_{H}=$ Horizontal component of earth's magnetic field $=3.6 \times 10^{-}$ ${ }^{5}$ Tesla.)
|
(a) 9.7, (b) 4.9, (c) 19.4, (d) 14.6
|
(a) 9.7
|
$\left(\mu_{0} / 4 \pi\right)\left(M / r^{3}\right)=3.6 \times 10^{-5}$ $M=\left[\left(3.6 \times 10^{-5}\right) /\left(10^{-7}\right)\right](0.3)^{3}$ $M=9.7 \mathrm{Am}^{2}$
|
Magnetism
|
Needles $\mathrm{N}_{1}, \mathrm{~N}_{2}$ and $\mathrm{N}_{3}$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will
|
(a) attract all three of them, (b) attract $\mathrm{N}_{1}$ and $\mathrm{N}_{2}$ strongly but repel $\mathrm{N}_{3}$, (c) attract $\mathrm{N}_{1}$ strongly, $\mathrm{N}_{2}$ weakly and repel $\mathrm{N}_{3}$ weakly, (d) attract $\mathrm{N}_{1}$ strongly, but repel $\mathrm{N}_{2}$ and $\mathrm{N}_{3}$ weakly
|
(c) Magnet will attract $\mathrm{N}_{1}$ strongly, $\mathrm{N}_{2}$ weakly and repel $\mathrm{N}_{3}$ weakly
|
Magnetism
|
|
A magnetic needle is kept in a non-uniform magnetic field. It experiences
|
(a) a force and a torque, (b) a force but not a torque, (c) a torque but not a force, (d) neither a force nor a torque
|
(a) A force and a torque act on a magnetic needle kept in a non-uniform magnetic field
|
Magnetism
|
|
The magnetic lines of force inside a bar magnet
|
(a) are from north-pole to south-pole of the magnet, (b) do not exist, (c) depend upon the area of cross-section of the bar magnet, (d) are from south-pole to north-pole of the magnet
|
(b) Materials of low retentivity and low coercivity are suitable for making electromagnets
|
Magnetism
|
|
Curie temperature is the temperature above which
|
(a) a ferromagnetic material becomes paramagnetic, (b) a paramagnetic material becomes diamagnetic, (c) a ferromagnetic material becomes diamagnetic, (d) a paramagnetic material becomes ferromagnetic
|
(a) a ferromagnetic material becomes paramagnetic
|
Magnetism
|
|
The materials suitable for making electromagnets should have
|
(a) high retentivity and high coercivity, (b) low retentivity and low coercivity, (c) high retentivity and low coercivity, (d) low retentivity and high coercivity
|
(b) Materials of low retentivity and low coercivity are suitable for making electromagnets.
|
Magnetism
|
|
Relative permittivity and permeability of a material are $\epsilon_{r}$ and $\mu_{r}$, respectively. Which of the following values of these quantities are allowed for a diamagnetic material?
|
(a) $\epsilon_{r}=1.5, \mu_{r}=1.5$, (b) $\epsilon_{r}=0.5, \mu_{r}=1.5$, (c) $\epsilon_{r}=1.5, \mu_{r}=0.5$, (d) $\epsilon_{r}=0.5, \mu_{r}=0.5$
|
(c) $\epsilon_{r}=1.5, \mu_{r}=0.5$
|
The values of relative permeability of diamagnetic materials are slightly less than 1 and $\varepsilon_{r}$ is quite high. Therefore $\varepsilon_{\mathrm{r}}=1.5$ and $\mu_{\mathrm{r}}=0.5$ could be the allowed values
|
Magnetism
|
In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a
|
(a) circle, (b) helix, (c) straight line, (d) ellipse
|
(c) straight line
|
Magnetic field exerts a force $=$ Bevsin0 $=0$ Electric field exerts force along a straight line The path of the charged particle will be a straight line
|
Magnetism
|
A charged particle moves through a magnetic field perpendicular to its direction. Then
|
(a) kinetic energy changes but the momentum is constant, (b) the momentum changes but the kinetic energy is constant, (c) both momentum and kinetic energy of the particle are not constant, (d) both momentum and kinetic energy of the particle are constant.
|
(b) the momentum changes but the kinetic energy is constant
|
The magnetic force will act perpendicular to velocity at every instant, the path will be circular. Due to change in direction momentum will change but the total energy will remain the same.
|
Magnetism
|
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to
|
(a) induction of electrical charge on the plate, (b) shielding of magnetic lines of force as aluminium is a paramagnetic material, (c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping, (d) development of air current when the plate is placed
|
(c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
|
Magnetism
|
|
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8: 27 . The ratio of the radii of the nuclei (assumed to be spherical) is
|
(a) $3: 2$
(b) $2: 3$
(c) 4: 9
(d) 8: 27
|
(a) 3: 2
|
$\operatorname{As}\left(v_{1} / v_{2}\right)=8 / 27 ;\left(r_{1} / r_{2}\right)=$ ?
Using law of conservation of linear momentum, $0=m_{1} \mathrm{~V}_{1}-\mathrm{m}_{2} \mathrm{~V}_{2}$
(As both are moving in opposite directions)
Or $m_{1} / m_{2}=v_{2} / v_{1}=27 / 8$
$$
\frac{\rho\left(\frac{4}{3}\right) \pi r_{1}^{3}}{\rho\left(\frac{4}{3}\right) \pi r_{2}^{3}}=\frac{27}{8}
$$
Therefore, $\left(r_{1} / r_{2}\right)=3 / 2$
|
Nuclear Physics
|
According to Bohr's theory, the time-averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the $\mathrm{n}^{\text {th }}$ orbit is proportional to ( $\mathrm{n}=$ principal quantum number)
|
(a) $\mathrm{n}^{-2}$
(b) $\mathrm{n}^{-3}$
(c) $\mathrm{n}^{-4}$
(d) $n^{-5}$
|
(d) $\mathrm{n}^{-5}$
|
Magnetic field at the centre, $B_{n}=\mu_{0} / / 2 r_{n}$
For a hydrogen atom, the radius of $\mathrm{n}^{\text {th }}$ orbit is given by
$r_{n}=\left(n^{2} / m\right)(h / 2 \pi)^{2}\left(4 \pi \varepsilon_{0} / e^{2}\right)$
$r_{n} \propto n^{2}$
$\mathrm{I}=\mathrm{e} / \mathrm{T}=\mathrm{e} /\left(2 \pi r_{n} / v_{n}\right)=e v_{n} / 2 \pi r_{n}$
Also, $\mathrm{v}_{\mathrm{n}} \propto \mathrm{n}^{-1} \therefore 1 \propto \mathrm{n}^{-3}$
Hence, $B_{n} \propto n^{-5}$
|
Nuclear Physics
|
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is (given binding energy per nucleon for deuteron $=1.1 \mathrm{MeV}$ and for helium $=7.0 \mathrm{MeV}$ )
|
(a) $25.8 \mathrm{MeV}$
(b) $32.4 \mathrm{MeV}$
(c) $30.2 \mathrm{MeV}$
(d) $23.6 \mathrm{MeV}$
|
(d) 23.6 MeV
|
${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}$
Energy released $=4\left(\right.$ Binding Energy of $\left.\left(1_{1}^{2} \mathrm{H}\right)\right)-4\left(\right.$ Binding Energy of $\left.\left(2{ }^{4} \mathrm{He}\right)\right)=4 \times 7-4 \times 1.1=23.6 \mathrm{MeV}$
|
Nuclear Physics
|
As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion
|
(a) kinetic energy decreases, potential energy increases but the total energy remains the same
(b) kinetic energy and total energy decrease but potential energy increases
(c) its kinetic energy increases but potential energy and total energy decrease
(d) kinetic energy, potential energy and total energy decrease.
|
(c) its kinetic energy increases but potential energy and total energy decrease
|
For an electron in $\mathrm{n}^{\text {th }}$ excited state of hydrogen atom,
kinetic energy $=\mathrm{e}^{2} /\left(8 \pi \epsilon_{0} \mathrm{n}^{2} \mathrm{a}_{0}\right)$
potential energy $=-e^{2} /\left(4 \pi \epsilon_{0} n^{2} a_{0}\right)$ and total energy $=-e^{2} /\left(8 \pi \epsilon_{0} n^{2} a_{0}\right)$
where ao is Bohr radius.
As an electron makes a transition from an excited state to the ground state, $\mathrm{n}$ decreases. Therefore kinetic energy increases but potential energy and total energy decrease.
|
Nuclear Physics
|
Energy required for the electron excitation in $\mathrm{Li}^{++}$from the first to the third Bohr orbit is
|
(a) $12.1 \mathrm{eV}$
(b) $36.3 \mathrm{eV}$
(c) $108.8 \mathrm{eV}$
(d) $122.4 \mathrm{eV}$
|
(c) $108.8 \mathrm{eV}$
|
Using, $E_{n}=\left(13.6 Z^{2} / n^{2}\right) \mathrm{eV}$
Here, $\mathrm{Z}=3$ (For $\mathrm{Li}^{++}$)
Therefore, $\mathrm{E}_{1}=\left(13.6(3)^{2} / 1^{2}\right)=-122.4 \mathrm{eV}$
and $E_{3}=\left(13.6(3)^{2} / 3^{2}\right)=-13.6 \mathrm{eV}$
$\Delta \mathrm{E}=\mathrm{E}_{3}-\mathrm{E}_{1}=-13.6+122.4=108.8 \mathrm{eV}$
|
Nuclear Physics
|
If the binding energy per nucleon in ${ }_{3} \mathrm{Li}^{7}$ and ${ }_{2} \mathrm{He}^{4}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$ respectively, then in the reaction
|
(a) $39.2 \mathrm{MeV}$
(b) $28.24 \mathrm{MeV}$
(c) $17.28 \mathrm{MeV}$
(d) $1.46 \mathrm{MeV}$
|
(c) 17.28 MeV
|
Binding energy of $3 \mathrm{Li}^{7}=7 \times 5.60=39.2 \mathrm{MeV}$
Binding energy of ${ }_{2} \mathrm{He}^{4}=4 \times 7.06=28.24 \mathrm{MeV}$
Energy of proton $=$ Energy of $\left[2\left(2 \mathrm{He}^{4}\right)-3 \mathrm{Li}^{7}\right]$
$=2 \times 28.24-39.2=17.28 \mathrm{MeV}$
|
Nuclear Physics
|
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be
|
(a) $2^{1 / 3}: 1$
(b) $1: 3^{1 / 2}$
(c) $3^{1 / 2}: 1$
(d) $1: 2^{1 / 3}$
|
(d) $1: 2^{1 / 3}$
|
Momentum is conserved during disintegration $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}--$-(1)
For an atom $R=R_{0} A^{1 / 3}$
$\left(R_{1} / R_{2}\right)=\left(A_{1} / A_{2}\right)^{1 / 3}=\left(m_{1} / m_{2}\right)^{1 / 3}=\left(v_{2} / v_{1}\right)^{1 / 3}$ from (1)
$\left(R_{1} / R_{2}\right)=(1 / 2)^{1 / 3}=1 / 2^{1 / 3}$
|
Nuclear Physics
|
A nuclear transformation is denoted by $X(n, \alpha)_{3} L i^{7}$. Which of the following is the nucleus of element $X$ ?
|
(a) ${ }_{5} \mathrm{~B}^{9}$
(b) ${ }_{4} \mathrm{Be}^{11}$
(c) ${ }_{6} \mathrm{C}^{12}$
(d) ${ }_{5} \mathrm{~B}^{10}$
|
(d) $5^{10}$
|
The nuclear transformation is given by
${ }_{2} X^{A}+{ }_{o n}{ }^{1} \rightarrow{ }_{2} \mathrm{He}^{4}+{ }_{3} \mathrm{Li}^{7}$
According to conservation of mass number
$A+1=4+7$ or $A=10$
According to conservation of charge number
$Z+0 \rightarrow 2+3$ or $Z=5$
So the nucleus of the element be ${ }_{5} \mathrm{~B}^{10}$.
|
Nuclear Physics
|
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At $t=0$ it was 1600 counts per second and at $t=8$ seconds it was 100 counts per second. The count rate observed, as counts per second, at $t=6$ seconds is close to
|
(a) 200
(b) 360
(c) 150
(d) 400
|
(a) 200
|
According to law of radioactivity, the count rate at $\mathrm{t}=8$ seconds is
$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
$\mathrm{dN} / \mathrm{dt}=\lambda N=\lambda N_{0} \mathrm{e}^{-\lambda t}$
$1600=\lambda N_{0} e^{0}=\lambda N_{0} \Rightarrow 100=\lambda N_{0} e^{-8 \lambda}=1600 e^{-8 \lambda}$
At $t=6 \mathrm{sec}$
$d N / d t=\lambda N_{0} e^{-6 \lambda}=1600 \times\left(e^{-2 \lambda}\right)^{3}=1600 \times(18)=200$
|
Nuclear Physics
|
Heat cannot by itself flow from a body at a lower temperature to a body at a higher temperature is a statement or consequence of
|
(a) The second law of thermodynamics
(b) conservation of momentum
(c) conservation of mass
(d) The first law of thermodynamics
|
Second law of thermodynamics.
|
Thermodynamics
|
|
Which of the following is incorrect regarding the first law of thermodynamics?
|
(a) It introduces the concept of internal energy
(b) It introduces the concept of entropy
(c) It is not applicable to any cyclic process
(d) It is a restatement of the principle of conservation of energy
|
Statements (b) and (c) are incorrect regarding the first law of thermodynamics.
|
Thermodynamics
|
|
Which of the following statements is correct for any thermodynamic system?
|
(a) The internal energy changes in all processes
(b) Internal energy and entropy are state functions
(c) The change in entropy can never be zero
(d) The work done in an adiabatic process is always zero
|
Internal energy and entropy are state functions
|
Thermodynamics
|
|
Which of the following parameters does not characterize the thermodynamic state of matter?
|
(a) temperature
(b) pressure
(c) work
(d) volume
|
The work does not characterize the thermodynamic state of matter
|
Thermodynamics
|
|
Even a Carnot engine cannot give 100 % efficiency because we cannot
|
(a) prevent radiation
(b) find ideal sources
(c) reach absolute zero temperature
(d) eliminate friction.
|
We cannot reach absolute zero temperature
|
Thermodynamics
|
|
Which statement is incorrect?
|
(a) All reversible cycles have the same efficiency
(b) The reversible cycle has more efficiency than an irreversible one
(c) Carnot cycle is a reversible one
(d) Carnot cycle has the maximum efficiency in all cycles
|
All reversible cycles do not have the same efficiency.
|
Thermodynamics
|
|
A Carnot engine operating between temperatures T1 and T2 has efficiency 1 / 6 When T2 is lowered by 62 K, its efficiency increases to 1 / 3. Then T1 and T2 are, respectively
|
(a) 372 K and 310 K
(b) 372 K and 330 K
(c) 330 K and 268 K
(d) 310 K and 248 K
|
372 K and 310 K
|
The efficiency of Carnot engine, η=1−(T2/T1) η=1/6 T2/T1=5/6 T1=6 T2/5 As per the question, when T2 is lowered by 62 K, then its efficiency becomes 1/3 1/3=[1−(T2−62/T1)] [T2−62/T1]=1−(1/3) (Using equa (1)) 5(T2−62)/6 T2=2/3 5 T2−310=4 T2 ⇒ T2=310 K From equation (1) T1=(6×310)/5=372 K
|
Thermodynamics
|
100 g of water is heated from 30∘ C to 50∘ C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 Jkg−1 K−1 )
|
(a) 4.2 kJ
(b) 8.4 kJ
(c) 84 kJ
(d) 2.1 kJ
|
8.4 kJ
|
Δ Q=ms Δ T Here m=100 g=100 ×10−3 Kg S=4184 J kg−1 K−1 and Δ T=(50−30)=20∘ C Δ Q=100 ×10−3 ×4184 ×20=8.4 ×103 J Δ Q=Δ U+Δ W Change in internal energy Δ U=Δ Q=8.4 ×103 J=8.4 kJ
|
Thermodynamics
|
200 g water is heated from 40∘ C to 60∘ C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water =4184 J / kg / K )
|
(a) 167.4 kJ
(b) 8.4 kJ
(c) 4.2 kJ
(d) 16.7 kJ
|
16.7 kJ
|
For isobaric process, Δ U=Q=ms Δ T Here, m=200 g=0.2 Kg, s=4184 J / Kg / K Δ T=60∘ C−40∘ C=20∘ C Δ U=(0.2)(4184)(20)=16736 J=16.7 kJ
|
Thermodynamics
|
The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7∘ C. The gas is (R=8.3 J mol−1 K−1)
|
(a) monoatomic
(b) diatomic
(c) triatomic
(d) a mixture of monoatomic and diatomic
|
diatomic
|
According to the first law of thermodynamics Δ Q=Δ U+Δ W For an adiabatic process, Δ Q=0 0=Δ U+Δ W Δ U=−Δ W nC v Δ T=−Δ W C v=−Δ W /n Δ T=−(−146)×103 /[(1 ×103) ×7]=20.8 Jmol−1 K−1 For diatomic gas, C v=(5/2) R=(5/2) ×8.3=20.8 Jmol−1 K−1 Hence, the gas is diatomic
|
Thermodynamics
|
From the following statements, concerning ideal gas at any given temperature T, select the correct.
|
(a)The coefficient of volume expansion at constant pressure is the same for all ideal gases
(b)The average translational kinetic energy per molecule of oxygen gas is 3 kT, k being Boltzmann constant
(c)The mean-free path of molecules increases with an increase in the pressure
(d)In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different
|
The coefficient of volume expansion at constant pressure is the same for all ideal gases
|
Y=d V /(V 0 ×d T) at a constant temperature Y=1/V 0(d V/d T) p since P V=R T PdV=RdT or (dV/dT)=R/P 0 Therefore, ϕ=(1/V 0)(R/P 0)=R/R T 0 γ=1/T 0 γ=1/273
|
Thermodynamics
|
Calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 ∘ C and it is defined under which of the following conditions?
|
(a) From 14.5∘ C to 15.5∘ C at 760 mm of Hg
(b) From 98.5∘ C to 99.5∘ C at 760 mm of Hg
(c) From 13.5∘ C to 14.5∘ C at 76 mm of Hg
(d) From 3.5∘ C to 4.5∘ C at 76 mm of Hg
|
From 14.5∘ C to 15.5∘ C at 760 mm of Hg
|
1 calorie is the amount of heat required to raise the temperature of 1 gm of water from 14.5∘ C to 15.5 ∘ C at 760 mm of Hg
|
Thermodynamics
|
The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N 2 (molar mass 28) molecules in eV at the same temperature is
|
(a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
|
0.048
|
Average Kinetic Energy =(3/2) KT It depends on temperature and does not depend on molar mass For both the gases, average translational kinetic energy will be the same ie., 0.048 eV
|
Thermodynamics
|
One mole of a monoatomic gas is heated at a constant pressure of 1 atmosphere from 0 K to 100 K. If the gas constant R=8.32 J / mol K, the change in internal energy of the gas is approximately [1998]
|
(a) 2.3 J
(b) 46 J
(c) 8.67 ×103 J
(d) 1.25 ×103 J
|
1.25 ×103 J
|
Δ U=C v dT=(3 R/2) Δ T Δ U=(3/2) ×(8.3) ×(100)=1.25 ×103 J
|
Thermodynamics
|
An ideal gas heat engine is operating between 227∘ C and 127∘ C. It absorbs 104 J of heat at the higher temperature. The amount of heat converted into work is
|
(a) 2000 J
(b) 4000 J
(c) 8000 J
(d) 5600 J
|
2000 J
|
η=1−(T 2/T 1) η=1−(127+273)/(227+273)=1−(400/500)=1/5 W=η Q 1=1/5×10 4=2000 J
|
Thermodynamics
|
Two masses of $1 \mathrm{~g}$ and $4 \mathrm{~g}$ are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is
|
(a) $4: 1$ (b) $\sqrt{ } 2: 1$ (c) $1: 2$ (d) $1: 16$
|
(c) $1: 2$
|
$p=\sqrt{2 E m}$ Where $\mathrm{p}$ denotes momentum, $\mathrm{E}$ denotes kinetic energy and $m$ denotes mass. $$ \frac{p_{1}}{p_{2}}=\sqrt{\frac{m_{1}}{m_{2}}} \frac{p_{1}}{p_{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2} $$
|
Work Power Energy
|
If a machine is lubricated with oil
|
(a) the mechanical advantage of the machine increases (b) the mechanical efficiency of the machine increases (c) both its mechanical advantage and efficiency increase (d) its efficiency increases, but its mechanical advantage decreases.
|
(b) When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.
|
When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.
|
Work Power Energy
|
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$ U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}} $$ between the atoms. if the dissociation energy of the molecule is $(U(x=\infty)-U$ at equilibrium), $D$ is
|
(a) $b^{2} / 6 a$ (b) $b^{2} / 2 a$ (c) $b^{2} / 12 a$ (d) $b^{2} / 4 a$
|
(d) $b^{2} / 4 a$
|
$\mathrm{U}(\mathrm{x}=\infty)=0$, at equilibrium $\mathrm{dU} / \mathrm{dx}=0$ $U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}}--(1)$ $-12 a x^{-13}+6 b x^{-7}=0$ $-12 a / x^{13}=6 b / x^{7}$ $x^{6}=2 a / b--(2)$ From 1 and 2 we have $$ U=\frac{a-b(2 a / b)}{(2 a / b)^{2}} $$ $U=-b^{2} / 4 a$ $D=\left(0+b^{2} / 4 a\right)$ $\mathrm{D}=\mathrm{b}^{2} / 4 \mathrm{a}$
|
Work Power Energy
|
At time $t=0$ s particle starts moving along the $x$-axis. If its kinetic energy increases uniformly with time ' $t$ ', the net force acting on it must be proportional to
|
(a) $\sqrt{ } \mathrm{t}$ (b) constant (c) $t$ (d) $1 / \sqrt{ } \mathrm{V}$
|
(d) $1 / \sqrt{t}$
|
Linear dependency with initial Kinetic Energy as zero is given as $\mathrm{KE}=\mathrm{kt}$, where $\mathrm{k}$ is the proportionality constant. Kinetic energy can be written as $\mathrm{KE}=1 / 2 \mathrm{mv}^{2}$ so we can write $1 / 2 m v^{2}=k t$ $$ v=\sqrt{\frac{2 k t}{m}} \frac{d v}{d t}=\sqrt{\frac{k}{2 m t}} $$ $F=m . d v / d t$ $$ F=m \sqrt{\frac{k}{2 m t}} $$ F $\infty 1 / \sqrt{t}$
|
Work Power Energy
|
This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. If two springs $S_{1}$ and $S_{2}$ of force constants $k_{1}$ and $k_{2}$, respectively, are stretched by the same force, it is found that more work is done on spring $\mathrm{S}_{1}$ than on spring $\mathrm{S}_{2}$. Statement-1: If stretched by the same amount, work done on $\mathrm{S}_{1}$, will be more than that on $\mathrm{S}_{2}$ Statement-2 $: k_{1}<k_{2}$
|
(a) Statement-1 is True, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1. (b) Statement-1 is False, Statement-2 is true (c) Statement-1 is True, Statement-2 is false (d) Statement-1 is True, Statement-2 is true and Statement-2 is the correct explanation of statement-1.
|
(b) Statement-1 is False, Statement-2 is true
|
Given same force $F=k_{1} x_{1}=k_{2} x_{2}$ $k_{1} / k_{2}=x_{1} / x_{2}$ $W_{1}=1 / 2 k_{1} x_{1}^{2}$ and $W_{2}=1 / 2 k_{2} x_{2}^{2}$ As $W_{1} / W_{2}>1$ so $1 / 2 k_{1} x_{1}{ }^{2} 11 / 2 k_{2} x_{2}{ }^{2}>1$ $\mathrm{Fx}_{1} / \mathrm{Fx}_{2}>1 \Rightarrow \mathrm{k}_{2} / \mathrm{k}_{1}>1$ Therefore $k_{2}>k_{1}$ Statement- 2 is true Or if $x_{1}=x_{2}=x$ $W_{1} / W_{2}=1 / 2 k_{1} x^{2} / 1 / 2 k_{2} x^{2}$ $W_{1} / W_{2}=k_{1} / k_{2}<1$\ $\mathrm{W}_{1}<\mathrm{W}_{2}$, Statement-1 is False
|
Work Power Energy
|
A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration ac is varying with time ' $t$ ' as ac $=k^{2} r t^{2}$ where ' $k$ ' is a constant. The power delivered to the particle by the force acting on it is
|
(a) $2 \pi \mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{~s}$ (b) $m^{2} r^{2} t$ (c) $\mathrm{mk}^{4} \mathrm{r}^{2} \mathrm{t} / 3$ (d) zero
|
(b) $\mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{t}$
|
Centripetal acceleration $\mathrm{a}_{\mathrm{c}}=\mathrm{k}^{2} \mathrm{rt}^{2}$ $\mathrm{V}^{2} / \mathrm{r}=\mathrm{k}^{2} \mathrm{t}^{2}$ $\mathrm{V}^{2}=\mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t}^{2}$ $\mathrm{mv}^{2} / 2=\mathrm{m} \mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t} 2 / 2$ Kinetic energy $=\mathrm{m}^{2} \mathrm{r}^{2} \mathrm{t}^{2} / 2$ $$ \frac{d}{d t}(K \cdot E)=m k^{2} r^{2} t $$ Power $=\mathrm{m} \mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t}$
|
Work Power Energy
|
A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is
|
(a) $\sqrt{u^{2}-2 g L}$ (b) $\sqrt{u^{2}-g L}$ (c) $\sqrt{2\left(u^{2}-g L\right)}$ (d) $\sqrt{2 g L}$
|
(c) $\sqrt{2\left(u^{2}-g L\right)}$
|
At lowest position B, Potential Energy $=0$ Kinetic Energy $=1 / 2$ mu Total Energy $=1 / 2 \mathrm{mu}$ At $\mathrm{C}$, when string is horizontal, Potential Energy $=\mathrm{mgL}$ Kinetic Energy $=1 / 2 \mathrm{mv}^{2}$ Total Energy $=1 / 2 \mathrm{mv}^{2}+\mathrm{mgL}$ Since energy is conserved, $1 / 2 m v^{2}+m g L=1 / 2 m u^{2}$ $v^{2}=u^{2}-2 g L$ Since $v$ is in vertical direction and $u$ is in horizontal direction, they are mutually perpendicular to each other. Change in velocity $=\sqrt{u^{2}+v^{2}}|\Delta \vec{v}|=$ $$ \sqrt{u^{2}+\left(u^{2}-2 g L\right)}|\Delta \vec{v}|=\sqrt{\left.2\left(u^{2}-g L\right)\right)} $$
|
Work Power Energy
|
When a rubber-band is stretched by a distance $x$, it exerts a restoring force of magnitude $F=a x+$ $b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is
|
(a) $a L^{2} / 2+b L^{3} / 3$ (b) $1 / 2\left(a L^{2} / 2+b L^{3} / 3\right)$ (c) $a L^{2}+b L^{3}$ (d) $1 / 2\left(a L^{2}+b L^{3}\right)$
|
(a) $a L^{2} / 2+b L^{3} / 3$
|
Work done by a variable force $$ W=\int \vec{F} \cdot \overrightarrow{d s} $$ Wherein $\operatorname{Vvec}\{\mathrm{F}\} F$ is the variable force and $\backslash \mathrm{vec}\{\mathrm{ds}\} d s$ is small displacement $F=a x+b x^{2}$ Work done in displacing rubber through $\mathrm{dx}=\mathrm{Fdx}$ $$ W=\int_{0}^{L}\left(a x+b x^{2}\right) d x $$ $\mathrm{W}=\mathrm{aL}^{2} / 2+\mathrm{bL} 3 / 3$
|
Work Power Energy
|
A person trying to lose weight by burning fat lifts a mass of $10 \mathrm{~kg}$ up to a height of $1 \mathrm{~m} 1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \mathrm{X}$ $10^{7} \mathrm{~J}$ of energy per kg which is converted to mechanical energy with a $20 \%$ efficiency rate. Take $\mathrm{g}=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$
|
(a) $12.89 \times 10^{-3} \mathrm{~kg}$ (b) $2.45 \times 10^{-3} \mathrm{~kg}$ (c) $6.45 \times 10^{3} \mathrm{~kg}$ (d) $9.89 \times 10^{-3} \mathrm{~kg}$
|
(a) $12.89 \times 10^{-3} \mathrm{Kg}$
|
Work done against gravity $=(\mathrm{mgh}) 1000$ in lifting 1000 times $=10 \times 9.8 \times 10^{3}$ $=9.8 \times 10^{4}$ Joule $20 \%$ of the efficiency is used to convert fat into energy $\left(20 \%\right.$ of $\left.3.8 \times 10^{7} \mathrm{~J}\right) \times \mathrm{m}=9.8 \times 10^{4}$ Where $m$ is mass $\mathrm{m}=12.89 \times 10^{-3} \mathrm{Kg}$
|
Work Power Energy
|
A body of mass $\mathrm{m}=10^{-2} \mathrm{~kg}$ is moving in a medium and experiences a frictional force $\mathrm{F}=-\mathrm{KV}^{2}$. Its initial speed is $v_{0}=10 \mathrm{~ms}^{-1}$. If, after $10 \mathrm{~s}$, its energy is, the value of $k$ will be
|
(a) $10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$ (b) $10^{-1} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$ (c) $10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$ (d) $10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$
|
(a) $10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$
|
$1 / 2 m v_{t}^{2}=1 / 8 m v_{0}{ }^{2}$ $V_{\mathrm{f}}=\mathrm{V}_{0} / 2=5 \mathrm{~m} / \mathrm{s}$ $V_{\mathrm{f}}=\mathrm{v}_{\mathrm{o}} / 2=5 \mathrm{~m} / \mathrm{s}$ $\left(10^{-2}\right) \mathrm{dV} / \mathrm{dt}=-\mathrm{kV}^{2}$ $\int_{10}^{5} \frac{d v}{v^{2}}=-100 k \int_{0}^{10} d t$ $1 / 5-1 / 10=100 k(10)$ $\mathrm{k}=10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$
|
Work Power Energy
|
A time-dependent force $F=6 t$ acts on a particle of mass $1 \mathrm{~kg}$. If the particle starts from rest, the work done by the force during the first $1 \mathrm{sec}$. will be :
|
(a) $9 \mathrm{~J}$ (b) $18 \mathrm{~J}$ (c) $4.5 \mathrm{~J}$ (d) $22 \mathrm{~J}$
|
(c) $4.5 \mathrm{~J}$
|
$F=6 t, m=1 \mathrm{Kg}, u=0$ Now, $F=m a=m(d v / d t)=1 x(d v / d t)$ $\mathrm{dv} / \mathrm{dt}=6 \mathrm{t}$ $\int_{0}^{v} d v=\int_{0}^{1} 6 t d t$ $v=3\left(1^{2}-0\right)=3 \mathrm{~m} / \mathrm{s}$ From work-energy theorem $W=\Delta K E=1 / 2 m\left(v^{2}-u^{2}\right)$ $W=1 / 2 \times 1\left(3^{2}-0\right)=4.5 \mathrm{~J}$
|
Work Power Energy
|
A particle is moving in a circular path of radius a under the action of an attractive potential $\mathrm{U}=$ $\mathrm{k} / 2 \mathrm{r}^{2}$.Its total energy is
|
(a) $k / 2 a^{2}$ (b) Zero (c) $-3 / 2\left(\mathrm{k} / \mathrm{a}^{2}\right)$ (d) $-k / 4 a^{2}$
|
(b) Zero
|
$\mathrm{F}=\mathrm{mv} v^{2}=\mathrm{K} / \mathrm{r}^{2}$ $K E=1 / 2 m v^{2}=K / 2 r^{2}$ Total energy $=$ P.E + K.E $-K / 2 r^{2}+K / 2 r^{2}=$ Zero
|
Work Power Energy
|
A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of
|
(a) $(2 / 3) \mathrm{k}$ (b) $(3 / 2) \mathrm{k}$ (c) $3 \mathrm{k}$ (d) $6 \mathrm{k}$
|
(b) (3/2) k
|
For a spring, $(K \times L)$ is a constant Therefore, $\mathrm{k} \times \mathrm{L}=\mathrm{k} \times \mathbf{x}(2 \mathrm{~L} / 3$ ) Where $2 L / 3=$ length of longer piece Or k'=(3/2)k
|
Work Power Energy
|
A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
|
(a) $v$ (b) $v^{2}$ (c) $v^{3}$ (d) $v^{4}$
|
(c) $v^{3}$
|
Force $=v \times d m / d t$ $$ \begin{aligned} & \text { Force }=\frac{d}{d t}(\text { volume } x \text { density })=v \frac{d}{d t}(\text { Axp }) \\ & \text { Force }=v \text { Ap } \frac{d x}{d t} \\ & \text { Power }=\text { Force } x \text { velocity } \\ & \left(\text { Apv }{ }^{2}\right) v=\mathrm{Apv}^{3} \\ & \text { Power } \propto \mathrm{V}^{3} \\ & \text { Answer: (c) } \mathbf{v}^{3} \end{aligned} $$
|
Work Power Energy
|
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
|
(a) 30 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
|
(c) 10 m/s
|
Just after collision
v_c = m1v1 + m2v2 / (m1 + m2)
v_c = (10 * 14 + 4 * 0) / (10 + 4)
v_c = 10 m/s
|
Centre of Mass
|
Two particles A and B, initially at rest, move towards each other under the mutual force of attraction. At the instant when the speed of A is v and the speed of B is 2v, the speed of the centre of mass of the system is
|
(a) 3v
(b) v
(c) 1.5v
(d) zero
|
(d) zero
|
FA is the force on particle A
FA = mA * aA = mA * v / t
FB is the force on particle B
FB = mB * aB = mB * 2v / t
Since FA = FB
mA * v / t = mB * 2v / t
So mA = 2mB
For the centre of mass of the system
v = (mA * vA + mB * vB) / (mA + mB)
v = (2mB * v - mB * 2v) / (2mB + mB) = 0
The negative sign is used because the particles are travelling in the opposite directions
|
Centre of Mass
|
A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 m/s. Which of the following statements(s) are correct for the system of these two masses
|
(a) The total momentum of the system is 3 kg m/s
(b) The momentum of 5 kg mass after the collision is 4 kg m/s
(c) The kinetic energy of the centre of mass is 0.75 J
(d) The total kinetic energy of the system is 4 J
|
(a) and (c)
|
If the velocity of 1 kg mass before the collision is u and velocity of 5 kg mass after the collision is v then
From the conservation of linear momentum
1 * u + 5 * 0 = 1 * (-2) + 5 * v
u = -2 + 5v
Since the collision is elastic, total kinetic energy will be equal
Therefore,
1/2 * 1 * u^2 = 1/2 * 1 * 2^2 + 1/2 * 5 * v^2
u^2 = 4 + 5v^2
(-2 + 5v)^2 = 4 + 5v^2
4 - 20v + 25v^2 = 0
-20v + 20v^2 = 0
-20v(1 - v) = 0
v = 1 m/s
Thus, u = 3 m/s
The velocity of the centres of mass of the combined system
V_cm = (m1v1 + m2v2) / (m1 + m2)
V_cm = (1 * (-2) + 5 * 1) / (1 + 5)
V_cm = 3 / 6
V_cm = 1 / 2
The combined Kinetic Energy of the system
Kinetic Energy = 1/2(1 + 5) * (1 / 2)^2
K.E = 6 / 8
Kinetic Energy = 0.75 J
Total momentum of the system
= (m1 + m2) * v_cm
= (1 + 5) * 1 / 2
= 3 kg-m/s
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Centre of Mass
|
Consider a rubber ball freely falling from a height h=4.9 m onto a horizontal elastic plate. Assume that the duration of a collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be
|
The velocity of the ball is reversed when it strikes the surface
|
The velocity of the ball is reversed when it strikes the surface
|
Centre of Mass
|
|
Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement-2: Principle of conservation of momentum holds true for all kinds of collisions.
|
(a) Statement-1 is true, Statement-2 is false
(b) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1
(c) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
(d) Statement-1 is false, Statement-2 is true
|
(b) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1
|
In a completely inelastic collision then m1 v1 + m2 v2 = m1 v + m2 v
v = (m1 v1 + m2 v2) / (m1 + m2)
Kinetic Energy = p1^2 / 2 m1 + p2^2 / 2 m2
As p1 and p2 both cannot be zero simultaneously therefore total kinetic energy cannot be lost.
|
Centre of Mass
|
A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collisions are perfectly inelastic, the percentage loss in the energy during the collision is close to
|
(a) 56%
(b) 62%
(c) 44%
(d) 50%
|
(a) 56%
|
Conservation of linear momentum in the x-direction
(pi)x = (pf)x or 2mv = (2m + m)Vx
vx = 2/3 V
Conservation of linear momentum in y-direction
(pi)y = (pf)y or 2mv = (2m + m) Vy
vy = 2/3 V
The initial kinetic energy of the two particles system is
Ei = 1/2 m (2v)^2 + 1/2 (2m) v^2
Ei = 1/2 * 4mv + 1/2 * 2mv^2
Ei = 2mv^2 + mv^2 = 3mv^2
Final energy of the combined two particles system is
Ef = 1/2 (3m) (vx^2 + vy^2)
Ef = 1/2 (3m) [4v^2 / 9 + 4v^2 / 9]
Ef = 4mv^2 / 3
Loss of energy ΔE = Ei - Ef
ΔE = mv^2 (3 - 4/3) = (5/3) mv^2
Percentage loss in the energy during the collision
ΔE / Ei * 100 = [(5/3) mv^2 / 3mv^2] * 100
= (5/9) * 100 ≈ 56%
|
Centre of Mass
|
The distance of the centre of mass of a solid uniform cone from its vertex is Z0. If the radius of its base is R and its height is h then Z0 is equal to
|
(a) 3h / 4
(b) h^2 / 4R
(c) 5h / 8
(d) 3h^2 / 8R
|
(a) 3h / 4
|
For a solid cone centre of mass from the base is h / 4
Position of centre of mass of a solid cone from the vertex = h - h / 4 = 3h / 4
|
Centre of Mass
|
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