Solid and Fluids - NEET Physics Chapterwise MCQs & PYQs

NEET Solid and Fluids MCQs & PYQs

Practice NEET Solid and Fluids Questions

Question 11:

easy

Two objects A & B of equal density and radius rA = 1 mm and rB = 2 mm are moving in same medium then find the ratio of their terminal velocity \( \frac{v_{B}}{v_{A}}\) in the medium.

The terminal velocity  v  of a spherical object moving through a viscous medium is directly proportional to the square of its radius, expressed as  v     

\(r^2 \), when the density and the medium remain constant.

\( \frac{v_B}{v_A} = \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 \)

Thus, the required ratio \( \frac{v_B}{v_A} \) is 4 (Option 3).

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Question 12:

easy

The viscous drag acting on a metal sphere of diameter \(1\text{ mm}\), falling through a fluid of viscosity \(0.8\text{ Pa s}\) with a velocity of \(2\text{ m s}^{-1}\) is equal to

According to Stokes' law, \(F = 6\pi \eta r v\). Here \(r = 0.5 \times 10^{-3}\text{ m}\), \(\eta = 0.8\text{ Pa s}\), and \(v = 2\text{ m s}^{-1}\). Calculating, we get \( F = 6 \times 3.14 \times 0.8 \times 0.5 \times 10^{-3} \times 2 \approx 15 \times 10^{-3}\text{ N}\).

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Question 13:

easy

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length \(100\text{ cm}\) to stretch it by \(1\text{ mm}\) is (if Young’s modulus of the wire \(= 2.0 \times 10^{11}\text{ N m}^{-2}\) )

Energy per unit volume is \(u = \frac{1}{2} \times Y \times (\text{strain})^2\). Strain \(= \frac{Delta l}{l} = \frac{10^{-3}\text{ m}}{1\text{ m}} = 10^{-3}\). Thus, \(u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (10^{-3})^2 = 10^5\text{ J/m}^3\).

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Question 14:

easy

If surface tension of a liquid is \(T\), then excess force required to lift a ring of radius \(R\) from surface of this liquid is:

A ring has two free surfaces in contact with the liquid (inner and outer boundaries). Therefore, the total contact length is \(L = 2(2\pi R) = 4\pi R\), and the excess force required is \(F = T \times 4\pi R\).

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Question 15:

easy

A drop of water falls in air with terminal velocity \(v_0\). If 27 such drops combine to form a big drop. Then terminal velocity of this new big drop will be:

When 27 identical small drops of radius \(r\) coalesce, the radius \(R\) of the single large drop is \(R = 27^{1/3}r = 3r\). Since terminal velocity \(v \propto r^2\), the new terminal velocity is \(v' = 3^2 v_0 = 9v_0\).

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Question 16:

easy

Given below are two statements:

Assertion (A): A hydrogen-filled balloon stops rising after it has attained a certain height in the sky.

Reason (R): The atmospheric pressure decreases with height and becomes zero when maximum height is attained by balloon.

 

As the balloon rises, the density of air decreases, leading to a decrease in buoyant force until it equals the weight of the balloon, so it stops rising. Thus, Assertion is true. However, atmospheric pressure does not become zero at this height, making Reason false.

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Question 17:

easy

Given below are two statements:

Assertion (A): When two soap bubbles of different radii are brought into contact, the common interface of contact bulges into the bubble of larger radii.

Reason (R): Pressure inside a soap bubble of a lesser radius is more than pressure inside the soap bubble of a larger radius.

 

The excess pressure inside a soap bubble is \(P = \frac{4T}{R}\). Thus, pressure is greater inside the smaller bubble (lesser radius). When in contact, the common interface bulges towards the larger bubble due to this pressure difference. Both are true and Reason explains Assertion.

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Question 18:

easy

Given below are two statements:

Assertion (A): Young’s modulus for a perfectly plastic body is zero.

Reason (R): For a perfectly plastic body, restoring force is zero.

 

For a perfectly plastic body, there is no tendency to regain its shape, meaning the restoring force (and thus stress) is zero. Since Young's modulus \(Y = \frac{\text{stress}}{\text{strain}}\), it is also zero. Both statements are true and Reason is the correct explanation.

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Question 19:

easy

A small ball of density \(rho\) is immersed in a liquid of density \(sigma (\sigma > \rho)\) to a depth h and released. The height above the surface of water upto which the ball jumps is:

Applying the work-energy theorem, work done by the buoyant force over depth \(h\) equals total work done against gravity: \(V \sigma g h = V \rho g (h + H)\). Solving for height \(H\) gives \(H = \left(\frac{\sigma}{\rho} - 1\right)h\).

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Question 20:

easy

An object of mass m and density \(\rho\) is falling in a viscous liquid of density \(\frac{\rho}{4}\). When it attains terminal velocity, then viscous force acting on it will be:

At terminal velocity, the upward forces balance the downward force: \(F_v + F_b = mg\). Since buoyant force is \(F_b = V \rho_l g = \left(\frac{m}{\rho}\right)\left(\frac{\rho}{4}\right)g = \frac{mg}{4}\), we get \(F_v = mg - \frac{mg}{4} = \frac{3mg}{4}\).

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