Mechanics

101. A mass is attached to a spring that hangs vertically. The extension pr

A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be :

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6 cm
1 cm
0 cm
36 cm
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According to Hooke’s Law, the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. The force stretching the vertical spring is the weight of the attached mass. The weight of the mass is proportional to the acceleration due to gravity.
– Hooke’s Law: Force (F) = spring constant (k) × extension (x), or F = kx.
– Weight (W) = mass (m) × acceleration due to gravity (g).
– On Earth: $W_{Earth} = mg_{Earth} = kx_{Earth}$.
– On the Moon: $W_{Moon} = mg_{Moon} = kx_{Moon}$.
Given: Extension on Earth $x_{Earth} = 6$ cm.
Acceleration due to gravity on the Moon $g_{Moon} = \frac{1}{6} g_{Earth}$.
From Hooke’s Law on Earth: $mg_{Earth} = k \times 6$ cm. So, $\frac{mg_{Earth}}{k} = 6$ cm.
On the Moon: $mg_{Moon} = kx_{Moon}$.
Substitute $g_{Moon}$: $m \left(\frac{g_{Earth}}{6}\right) = kx_{Moon}$.
$\frac{1}{6} (mg_{Earth}) = kx_{Moon}$.
Substitute $mg_{Earth} = k \times 6$ cm: $\frac{1}{6} (k \times 6 \text{ cm}) = kx_{Moon}$.
$k \times 1 \text{ cm} = kx_{Moon}$.
$x_{Moon} = 1$ cm.
The extension is directly proportional to the weight, and thus directly proportional to gravity. Since gravity on the Moon is one-sixth of Earth’s, the extension will also be one-sixth.

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102. One block of 2·0 kg mass is placed on top of another block of 3·0 kg m

One block of 2·0 kg mass is placed on top of another block of 3·0 kg mass. The coefficient of static friction between the two blocks is 0·2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s², the maximum value of the frictional force is :

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50 N
30 N
4 N
10 N
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When the two blocks move together without slipping, the force responsible for the acceleration of the top block (mass 2.0 kg) is the static friction force exerted by the bottom block on the top block. The maximum value this static friction force can attain is given by the product of the coefficient of static friction and the normal force acting on the top block.
– The normal force on the top block is equal to its weight: N = m₁g.
– The maximum static friction force is given by $f_{s,max} = \mu_s N$.
– For the blocks to move together without slipping, the actual static friction force must be less than or equal to $f_{s,max}$. The question asks for the maximum value of the frictional force in this scenario, which is the maximum static friction force that can exist between the blocks.
Given: mass of top block m₁ = 2.0 kg, mass of bottom block m₂ = 3.0 kg, coefficient of static friction $\mu_s = 0.2$, acceleration due to gravity g = 10 m/s².
Normal force on the top block (due to the bottom block) N = m₁g = 2.0 kg × 10 m/s² = 20 N.
Maximum static friction force $f_{s,max} = \mu_s N = 0.2 \times 20 N = 4 N$.
For the blocks to move together, the actual friction force on the top block is $f = m_1 a$, where ‘a’ is the acceleration of the system. As the pulling force F on the bottom block increases, ‘a’ increases, and thus the required friction force ‘f’ increases. Slipping occurs when the required ‘f’ exceeds $f_{s,max}$. The maximum value the frictional force can reach while preventing slipping is $f_{s,max}$.

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103. The power required to lift a mass of 8·0 kg up a vertical distance of

The power required to lift a mass of 8·0 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m/s²):

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80 W
160 W
320 W
640 W
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The power required to lift an object is equal to the work done in lifting the object divided by the time taken. Work done in lifting is equal to the force required to lift it multiplied by the vertical distance lifted. The minimum force required to lift a mass vertically at a constant speed or to overcome gravity is equal to its weight.
– Weight (Force) = mass × acceleration due to gravity (W = mg).
– Work done (W) = Force × distance (W = Fd).
– Power (P) = Work done / Time (P = W/t).
Given: mass m = 8.0 kg, distance d = 4 m, time t = 2 s, acceleration due to gravity g = 10 m/s².
Force required to lift the mass = W = mg = 8.0 kg × 10 m/s² = 80 N.
Work done = Fd = 80 N × 4 m = 320 Joules.
Power = Work done / Time = 320 J / 2 s = 160 Watts.

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104. Sand falls vertically on a conveyor belt at a rate of 0·1 kg/s. In ord

Sand falls vertically on a conveyor belt at a rate of 0·1 kg/s. In order to keep the belt moving at a uniform speed of 2 m/s, the force required to be applied on the belt is :

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0 N
0·2 N
1·0 N
2·0 N
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To keep the conveyor belt moving at a uniform speed, a force must be applied to counteract the change in momentum of the sand as it lands on the belt and accelerates to the belt’s speed. Sand falls vertically with zero initial horizontal momentum. As it lands on the belt, it acquires the belt’s horizontal velocity of 2 m/s. The rate at which mass is added to the belt is dm/dt = 0.1 kg/s. The force required is equal to the rate of change of momentum in the horizontal direction. The momentum added per unit time is (dm/dt) * v, where v is the velocity of the belt. Force F = (dm/dt) * v = (0.1 kg/s) * (2 m/s) = 0.2 N. This force is needed to continuously accelerate the newly added sand horizontally from rest to the belt’s speed.
The force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt’s velocity.
This problem involves a variable mass system, specifically where mass is being added. The relevant physical principle is Newton’s second law in its momentum form: F_net = dP/dt. In this case, the external force applied to the belt is responsible for increasing the horizontal momentum of the sand falling onto it.

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105. A railway wagon (open at the top) of mass M₁ is moving with speed v₁ a

A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is :

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[amp_mcq option1=”v₁ = v₂” option2=”1/2 M₁v₁² < 1/2 M₂v₂²" option3="M₁v₁ = M₂v₂" option4="M₁v₁ < M₂v₂" correct="option3"]

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As the rain falls vertically, it has no horizontal momentum. The horizontal momentum of the system (wagon + water) is conserved because there are no external horizontal forces acting on it (assuming negligible friction and air resistance).
The initial horizontal momentum of the wagon is $P_1 = M_1 v_1$. After the rain has collected, the total mass is $M_2$ and the speed is $v_2$. The final horizontal momentum is $P_2 = M_2 v_2$. By conservation of horizontal momentum, $P_1 = P_2$, so $M_1 v_1 = M_2 v_2$.
This is an example of an inelastic collision/process where mass is added to the system. Kinetic energy is not conserved in this case; the kinetic energy of the system decreases because the incoming water had zero horizontal kinetic energy relative to the ground.

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106. Two forces of 5.0 N each are acting on a point mass. If the angle betw

Two forces of 5.0 N each are acting on a point mass. If the angle between the forces is 60°, then the net force acting on the point mass has magnitude close to :

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8.6 N
4.3 N
5.0 N
6.7 N
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The correct option is A.
The magnitude of the resultant of two vectors (forces) can be found using the formula derived from the law of cosines: $|R| = \sqrt{|F_1|^2 + |F_2|^2 + 2|F_1||F_2|\cos\theta}$, where $\theta$ is the angle between the forces.
Given $|F_1| = 5.0$ N, $|F_2| = 5.0$ N, and $\theta = 60°$. The magnitude of the resultant force is $|R| = \sqrt{5.0^2 + 5.0^2 + 2 \cdot 5.0 \cdot 5.0 \cdot \cos(60°)}$.
$|R| = \sqrt{25 + 25 + 2 \cdot 25 \cdot (1/2)} = \sqrt{50 + 25} = \sqrt{75}$.
$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$.
Using the approximation $\sqrt{3} \approx 1.732$, $|R| \approx 5 \times 1.732 = 8.66$.
The closest value among the options is 8.6 N.

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107. A particle is moving in a circle of radius R with a constant speed v.

A particle is moving in a circle of radius R with a constant speed v. Its average acceleration over the time when it moves over half the circle is :

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$ rac{v^2}{R}$
$ rac{pi v^2}{2R}$
$ rac{2v^2}{pi R}$
0
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The correct option is C.
Average acceleration is defined as the change in velocity divided by the time taken. For a particle moving in a circle with constant speed, the velocity vector changes direction continuously. Over half a circle, the initial and final velocity vectors are in opposite directions.
Let the particle start at $\theta=0$ (position vector $R\hat{i}$) and move counter-clockwise. The initial velocity is $\vec{v}_i = v\hat{j}$. After moving half a circle ($\pi$ radians), it is at $\theta=\pi$ (position vector $-R\hat{i}$). The final velocity is $\vec{v}_f = -v\hat{j}$. The change in velocity is $\Delta\vec{v} = \vec{v}_f – \vec{v}_i = -v\hat{j} – v\hat{j} = -2v\hat{j}$. The magnitude of the change in velocity is $|\Delta\vec{v}| = 2v$. The distance covered is half the circumference, $\pi R$. The time taken is $t = (\pi R) / v$. The average acceleration is $\vec{a}_{avg} = \Delta\vec{v}/t = (-2v\hat{j}) / (\pi R / v) = -\frac{2v^2}{\pi R}\hat{j}$. The magnitude of the average acceleration is $|\vec{a}_{avg}| = \frac{2v^2}{\pi R}$.

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108. Two identical containers X and Y are connected at the bottom by a thin

Two identical containers X and Y are connected at the bottom by a thin tube of negligible volume. The tube has a valve in it, as shown in the figure. Initially container X has a liquid filled up to height h in it and container Y is empty. When the valve is opened, both containers have equal amount of liquid in equilibrium. If the initial (before the valve is opened) potential energy of the liquid is Pᵢ and the final potential energy is Pғ, then:

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Pᵢ=Pғ
Pғ = 1/4 * Pᵢ
Pᵢ=2Pғ
Pғ = 1/8 * Pᵢ
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The correct option is C.
Potential energy of a continuous mass distribution like liquid is the integral of $y \cdot dm \cdot g$. For a liquid column of uniform density in a cylindrical container, the center of mass is at half the height. The potential energy is proportional to the mass and the height of the center of mass. When the liquid redistributes to reach equilibrium in connected identical containers, the total volume is conserved, and the liquid levels become equal.
Let A be the cross-sectional area of the identical containers. Initially, all liquid is in X up to height h. Volume $V = Ah$. Mass $m = \rho V = \rho Ah$. The initial potential energy is $P_i = mg(h/2) = (\rho Ah)g(h/2) = \frac{1}{2}\rho Ag h^2$. When the valve is opened, the liquid distributes equally by volume between X and Y, and the levels are equal. Let the final height in both be $h_f$. Total volume $Ah = A h_f + A h_f = 2Ah_f$, so $h_f = h/2$. The liquid in X has mass $m_X = \rho A (h/2)$, and its CM is at $(h/2)/2 = h/4$. PE in X is $P_{fX} = m_X g (h/4) = (\rho Ah/2)g(h/4) = \frac{1}{8}\rho Ag h^2$. Similarly, PE in Y is $P_{fY} = \frac{1}{8}\rho Ag h^2$. The total final potential energy is $P_f = P_{fX} + P_{fY} = \frac{1}{8}\rho Ag h^2 + \frac{1}{8}\rho Ag h^2 = \frac{1}{4}\rho Ag h^2$. Comparing $P_i$ and $P_f$: $P_i = \frac{1}{2}\rho Ag h^2$ and $P_f = \frac{1}{4}\rho Ag h^2$. Thus, $P_i = 2P_f$.

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109. A sphere of volume V is made of a material with lower density than wat

A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f₁V (f₁ < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f₂V. Then: [amp_mcq option1="f₂=f₁" option2="f₂ = (a/g) * f₁" option3="f₂ > f₁” option4=”f₂ = (g/(g-a)) * f₁” correct=”option1″]

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This question was previously asked in
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The correct option is A.
The fraction of volume submerged when an object floats in a fluid is determined by the ratio of the object’s density to the fluid’s density ($f = \rho_{object} / \rho_{fluid}$). This relationship arises from the condition that the buoyant force equals the weight of the object. Both the buoyant force (which depends on the effective acceleration) and the weight (which also depends on the effective acceleration) are directly proportional to the effective acceleration. Therefore, the ratio of densities, and hence the fraction of submerged volume, is independent of the magnitude of the uniform effective acceleration (like gravity) acting on the system.
On Earth, the buoyant force is $F_{B1} = \rho_w (f_1V) g$ and the weight is $W_1 = \rho_s V g$. Since it floats, $F_{B1} = W_1 \implies \rho_w f_1 V g = \rho_s V g \implies f_1 = \rho_s / \rho_w$. On the spaceship with effective acceleration $a$, the buoyant force is $F_{B2} = \rho_w (f_2V) a$ and the weight is $W_2 = \rho_s V a$. Since it floats, $F_{B2} = W_2 \implies \rho_w f_2 V a = \rho_s V a \implies f_2 = \rho_s / \rho_w$. Thus, $f_1 = f_2$. The magnitude of the effective acceleration affects the magnitude of the forces but not their ratio when determining the submerged fraction.

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110. Two identical spring balances S₁ and S₂ are connected one after the ot

Two identical spring balances S₁ and S₂ are connected one after the other and are held vertically as shown in the figure. A mass of 10 kg is hanging from S₂. If the readings on S₁ and S₂ are W₁ and W₂ respectively, then:

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W₁=5kg and W₂=10kg
W₁=10kg and W₂=5kg
W₁=5kg and W₂=5kg
W₁=10kg and W₂=10kg
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The correct option is D.
When identical spring balances are connected in series and a mass is suspended from the lower one, the lower balance measures the weight of the suspended mass. The upper balance measures the total tension required to support everything hanging below it, which includes the lower balance (assuming negligible weight for the balance itself) and the suspended mass. Thus, both balances will read the weight of the 10 kg mass.
Spring balances measure force (tension). When used to measure weight under gravity, they are often calibrated to display mass (mass = weight / g). In this setup, the tension in the lower spring (S₂) is equal to the weight of the 10 kg mass. The tension in the upper spring (S₁) is equal to the sum of the tensions from S₂ and the weight of S₂ itself. Assuming the spring balances have negligible weight, S₁ measures the same tension as S₂, which is the weight of the 10 kg mass. If the mass is 10 kg, its weight is approximately 10g N. A spring balance calibrated in kg would read 10 kg for a 10 kg mass under standard gravity.

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