Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Kinematics 2D
Q2: A particle P is sliding down a frictionless hemispherical bowl. It passes point A at t=0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t= 0 along the horizontal string AB, with speed v. Friction between the bead and the string may be neglected. Let t_p and t_q be the respective times taken by P and Q to reach point B.
Options: t_p < t_Q
t_p > t_Q
t_p = t_Q
t_p / t_Q = (length of arc ABC / length of chord AB )
Answer: (1) t_P < t_Q
Solution: Horizontal displacements of the particle (P) and bead (Q) are equal. Bead Q moves along AB with a constant velocity v. For particle P, motion between AC will be under acceleration while between CB will be under retardation. The horizontal component of its velocity may be greater than or equal to v. It will not be less than v. Obviously, t_P < t_Q.
Topic: Kinematics 2D
Q3: A train is moving along a straight line with a constant acceleration 'a'. A girl standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The girl has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s², is
Answer: a = 5 m/s²
Solution: R = 1 / 2 at² + 1.15
u² sin 2θ / g = 1 / 2 a (2 u sin θ / g)² + 1.15
100 × √3 / 2 / 10 = 1 / 2 a (4 × 100 × 3 / 4 / 100) + 1.15
5 √3 = 3 a / 2 + 1.15
2 × 5 √3 = 3 a + 2.30
10 √3 = 3 a + 2.30
10 √3 - 2.30 = 3 a
10 × 1.73 - 2.3 = 3 a
17.3 - 2.3 = 3 a
15 / 3 = a
a = 5 m/s²
Topic: Kinematics 2D
Q5: A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river in the shortest time. The velocity of river water in km/hr is
Options: 1
3
4
√41
Answer: (2) 3 km/h
Solution: Time t = 15 min = 15 / 60 = 1 / 4 hour
Velocity = distance / time
V_b cos θ = 1 / (1 / 4) = 4 km/h
5 cos θ = 4 km/h
cos θ = 4 / 5
sin θ = 3 / 5
Consider the triangle of velocity ABC,
V_r / V_b = 3 / 5
V_r / 5 = 3 / 5
Or V_r = 3 km/h
Topic: Kinematics 2D
Q8: A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in metres, is
Answer: H₂ = 30 m
Solution: H₁ = u² sin² 45 / 2g = 120
u² / 2g = 120
When half of the kinetic energy is lost v = u / √2
H₂ = (u / √2)² sin² 30 / 2g = u² / 16g
From (1) and (2)
H₂ = H₁ / 4 = 30 m
Topic: Kinematics 2D
Q10: Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance). Is the statement true or false?
Answer: True
Solution: $v^2=u^2-2gH$
As the body falls back to the point of projection, $H=0$
$$v^2=u^2-0=u^2$$
$$\operatorname{Or} v=u$$
Hence the statement is true
Topic: Kinematics 2D
Q11: A projectile fired from the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path. Is the statement true?
Answer: True
Solution: The horizontal velocity of the projectile $=$ $u \cos \theta$
It remains constant throughout the motion
The vertical velocity becomes zero at the top of the path
Therefore the velocity is minimum at the top
Hence the statement is true
Topic: Kinematics 2D
Q12: Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. Is the statement true?
Answer: False
Solution: One train is moving in the direction of the earth's rotation and the other moves in the opposite direction. Hence they will exert different pressures on the rails. The statement is false.
Topic: Kinematics 2D
Q13: A particle moves in a circle of radius R. In half the period of revolution its displacement is and distance covered is
Answer: Displacement $AB=2r$
Distance $ACB=2\pi r / 2=\pi r$
Solution: Displacement $AB=2r$
Distance $ACB=2\pi r / 2=\pi r$
Topic: Kinematics 2D
Q14: Four persons $K, L, M, N$ are initially at the four corners of a square of side $d$. Each person now moves with a uniform speed $V$ in such a way that $K$ always moves directly towards $L$, $L$ directly towards $M$, $M$ directly towards $N$, and $N$ directly towards $K$. The four persons will meet at a time
Answer: Time Taken $=(d / \sqrt{2}) / (v / \sqrt{2})=d / v$
Solution: It is inferred on the basis of symmetry that all four persons will meet at $O$, the centre of the square.
At any instant, the component velocity along
$$KO=v \cos 45^\circ=v / \sqrt{2}$$
Distance $KO=d \cos 45^\circ=d / \sqrt{2}$
Time taken $=$ Distance $KO$ / velocity along $KO$
Time Taken $=(d / \sqrt{2}) / (v / \sqrt{2})=d / v$
Topic: Kinematics 2D
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
Answer: Theta=\tan ^{-1} a$$
Solution: $y=ax-bx^{2}$
Comparing the given equation with
$$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$$
Topic: Current Electricity
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}$)
Options: 3
0.2
2
0.02
Answer: (d) 0.02
Solution: 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$
Topic: 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
Options: $240 \mathrm{~W}$
$120 \mathrm{~W}$
$60 \mathrm{~W}$
$30 \mathrm{~W}$
Answer: (a) $240 \mathrm{~W}$
Solution: 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}$
Topic: 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
Options: $11 \times 10^{-4} \mathrm{~W}$
$11 \times 10^{-5} \mathrm{~W}$
$11 \times 10^{5} \mathrm{~W}$
$11 \times 10^{-3} \mathrm{~W}$
Answer: (b) $11 \times 10^{-5} \mathrm{~W}$
Solution: 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}$
Topic: 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
Options: 490
495
395
480
Answer: (c) 395
Solution: 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$
Topic: 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
Options: $R=0.001 r$
$R=r$
$R=2 r$
$R=1000 r$
Answer: (b) $\mathrm{R}=\mathrm{r}$
Solution: 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$
Topic: 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
Options: $(7 / 2) \Omega$
$(5 / 2) \Omega$
$(12 / 5) \Omega$
$(5 / 3) \Omega$
Answer: (d) (5/3) $\Omega$
Solution: $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$
Topic: 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?
Options: 990
505
550
910
Answer: (c) 550
Solution: 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$
Topic: 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
Options: Increased 8 times
Unchanged
Doubled
Halved
Answer: (a) Increased 8 times
Solution: 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}$
Topic: 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?
Options: $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}$
Answer: (b) $2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
Solution: 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}$
Topic: 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
Options: $400 \Omega$
$100 \Omega$
$200 \Omega$
$1600 \Omega$
Answer: (d) $1600 \Omega$
Solution: 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$
Topic: 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
Options: $20 \mathrm{~mA}$
$0.4 \mathrm{~mA}$
$100 \mathrm{~mA}$
$63 \mathrm{~mA}$
Answer: (a) $20 \mathrm{~mA}$
Solution: 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}$$
Topic: 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
Options: $14 \mathrm{~A}$
$8 \mathrm{~A}$
$10 \mathrm{~A}$
$12 \mathrm{~A}$
Answer: (d) $12 \mathrm{~A}$
Solution: 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
$\mathrm{I}=\mathrm{P} / \mathrm{N}=2500 / 220=11.36 \mathrm{~A} \approx 12 \mathrm{~A}$
Topic: Current Electricity
Q13: If a wire is stretched to make it $0.1 \%$ longer, its resistance will
Options: increase by $0.05 \%$
increase by $0.2 \%$
decrease by $0.2 \%$
decrease by $0.05 \%$
Answer: (b) increase by $0.2 \%$
Solution: 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 $0.2 \%$
Topic: 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
Options: 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.
Answer: (a) flow from antimony to bismuth at the cold junction
Solution: Antimony-bismuth couple is $A B C$ couple. It means that current flows from $A$ to $B$ at a cold junction
Topic: 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
Options: 3 ohm
2 ohm
1 ohm
4 ohm
Answer: (d) 4 ohm
Solution: $R_{t}=R_{0}(1+\alpha t)$
$R_{t}$ 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
Topic: Kinematics 1D
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)
Options: (a) 10 m
(b) 30 m
(c) 20 m
(d) 40 m
Answer: (d) 40 m
Solution: 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
Topic: 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
Options: (a) 25 / 11
(b) 3 / 2
(c) 5 / 2
(d) 11 / 5
Answer: (d) 11 / 5
Solution: 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
Topic: 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
Options: (a) 7 units
(b) 8.5 units
(c) 10 units
(d) 7\sqrt{2} units
Answer: (d) 7\sqrt{2} units
Solution: 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}
Topic: 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)
Options: (a) 100 m
(b) 75 m
(c) 160 m
(d) 150 m
Answer: (c) 160 m
Solution: 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
Topic: 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?
Options: (a) 293 m
(b) 111 m
(c) 91 m
(d) 182 m
Answer: (a) 293 m
Solution: 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.
Topic: 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
Options: (a) s = 1 / 2 ft²
(b) s = (1 / 4) ft²
(c) s = ft
(d) s = (1 / 72) ft²
Answer: (d) s = (1 / 72) ft²
Solution: 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
Topic: Kinematics 1D
Q10: Which of the following statements is false for a particle moving in a circle with a constant angular speed?
Options: (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
Answer: (b) The acceleration vector acts along the radius of the circle. The given statement is false.
Topic: 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
Options: (a) $v_{B}>v_{A}$
(b) $v_{A}=v_{B}$
(c) $v_{A}>v_{B}$
(d) their velocities depend on their masses
Answer: (b) $v_{A}=v_{B}$
Solution: 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)
$v_{A}=v_{B}$
Topic: 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?
Options: (a) $1 \mathrm{~cm}$
(b) $2 \mathrm{~cm}$
(c) $3 \mathrm{~cm}$
(d) $4 \mathrm{~cm}$
Answer: (a) $1 \mathrm{~cm}$
Solution: 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)
$8 \mathrm{ax}=8 \mathrm{a} \quad \mathrm{x}=1 \mathrm{~cm}$
Topic: 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
Options: (a) $120 \mathrm{~s}$
(b) $15 \mathrm{~s}$
(c) [latex]2\textbackslash sqrt\{2\}[/latex] s
(d) $110 \mathrm{~s}$
Answer: (c) [latex]2\textbackslash sqrt\{2\}[/latex] s
Solution: 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., $\mathrm{t}^{2}=200$
$\mathrm{t}=[2\textbackslash sqrt{2}]$
Topic: 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
Options: (a) $g H=(n-2) u^{2}$
(b) $2 g H=n 2 u^{2}$
(c) $g H=(n-2) 2 u^{2}$
(d) $2 g H=n u^{2}(n-2)$
Answer: (d) $2 g H=n u^{2}(n-2)$
Topic: Kinetic Theory of Gases
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
Options: (a) 20 RT
(b) 12 RT
(c) 4 RT
(d) 15 RT
Answer: (d) 15 RT
Solution: $U=\left(f_{1} / 2\right) n_{1} R T+\left(f_{2} / 2\right) n_{2} R T$
$U=(5 / 2)(3 R T)+(3 / 2)(5 R T)$
$U=15 R T$