Solid and Fluids - NEET Physics Questions
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Solid and Fluids

Practice Questions:
Question 21: moderate

The reading of spring balance when a block is suspended from it in air, is 60 N. This reading is changed to 40 N when the block is immersed in water. The specific gravity of the block is :

1. 3
2. 2
3. 6
4. 3/2
View Answer

Solution:

  • Loss of weight in water = $$\text{Weight in air} - \text{Weight in water} = 60\text{ N} - 40\text{ N} = 20\text{ N}$$

  • Specific gravity = $$\frac{\text{Weight in air}}{\text{Loss of weight in water}} = \frac{60}{20} = 3$$

Therefore, the specific gravity of the block is 3 (Option 1).

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Question 22: moderate

An open U-tube contains mercury. When 11.2 cm of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial level ?

1. 0.82 cm
2. 1.35 cm
3. 0.41 cm
4. 2.32 cm
View Answer

When 11.2 cm of water is poured into one arm, it balances a mercury column of height 2x, where x is the height the mercury rises in the other arm.

Using the pressure balance equation (density of water * height of water = density of mercury * 2x), we get 1 * 11.2 = 13.6 * 2x. Solving for x yields 0.41 cm, making Option 3 the correct answer.

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Question 23: moderate

A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of
water flowing out per second from the holes are both same. Then, R is equal to:

1. \[\frac{L}{\sqrt{2\pi}}\]
2. \[2\pi L\]
3. L
4. \[\frac{L}{2\pi}\]
View Answer

According to Torricelli's Law, the velocity of efflux from a hole at a depth h is given by $$v = \sqrt{2gh}$$.

Since the volume flow rate per second

\( Q = \text{Area}\times\text{Velocity} \) is the same for both holes:

$$A_{\text{square}} \cdot v_1 = A_{\text{circle}} \cdot v_2$$
$$L^2 \sqrt{2gy} = (\pi R^2) \sqrt{2g(4y)}$$
$$L^2 = \pi R^2 \cdot 2$$
$$R = \frac{L}{\sqrt{2\pi}}$$

Thus, the correct option is Option 1.

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Question 24: moderate

A tank is filled to a height H. The range of water coming out of a hole which is a depth H/4 from the surface of water level is :

1. \[\frac{2H}{\sqrt{3}}\]
2. \[\frac{\sqrt{3}H}{2}\]
3. \[\sqrt{3}H\]
4. \[\frac{3H}{4}\]
View Answer

The horizontal range of water emerging from a hole is given by the formula

$$R = 2\sqrt{h(H - h)} $$

$$R = 2\sqrt{\left(\frac{H}{4}\right)\left(\frac{3H}{4}\right)} = 2\left(\frac{\sqrt{3}H}{4}\right) = \frac{\sqrt{3}H}{2}$$
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Question 25: moderate

A simple pendulum oscillating in air has a period of \(\sqrt{3}\text{ s}\). If it is completely immersed in non-viscous liquid, having density \((\frac{1}{4})^{\text{th}}\) of the material of the bob, the new period will be

1. 2 s
2. \(\frac{\sqrt{3}}{2}\text{ s}\)
3. \(2\sqrt{3}\text{ s}\)
4. \(\frac{2}{\sqrt{3}}\text{ s}\)
View Answer

The effective acceleration due to gravity in the liquid is \(g' = g\left(1 - \frac{\rho_L}{\rho_B}\right) = g\left(1 - \frac{1}{4}\right) = \frac{3}{4}g\). Since \(T \propto \frac{1}{\sqrt{g}}\), the new period is \(T' = T\sqrt{\frac{g}{g'}} = \sqrt{3}\sqrt{\frac{4}{3}} = 2\text{ s}\).

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Question 26: moderate

A uniform rope of density \(rho\) and length \(L\) is hanging from roof. If young’s modulus of material of rope is \(Y\), then elongation produced in rope due to its own weight is:

1. \(\frac{\rho gL}{2Y}\)
2. \(\frac{\rho gL^2}{2Y}\)
3. \(\frac{\rho gL^2}{2AY}\)
4. \(\frac{\rho gL^2}{Y}\)
View Answer

The elongation of a uniform rope under its own weight is given by \(\Delta L = \frac{MgL}{2AY}\). Substituting mass \(M = \rho A L\), we obtain \(\Delta L = \frac{\rho g L^2}{2Y}\).

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Question 27: moderate

A rubber sphere is taken in a lake to a depth \(1800\text{ m}\). If bulk modulus of rubber is \(6 \times 10^8\text{ N/m}^2\), then radius of this rubber sphere will decrease by:

1. 1%
2. 2%
3. 3%
4. 4%
View Answer

The pressure change is \(dP = \rho g h = 10^3 \times 10 \times 1800 = 1.8 \times 10^7\text{ N/m}^2\). The fractional volume change is \(\frac{dV}{V} = \frac{dP}{B} = \frac{1.8 \times 10^7}{6 \times 10^8} = 3\%\). Since \(\frac{dV}{V} = 3\frac{dr}{r}\), the radius decreases by \(\frac{3\%}{3} = 1\%\).

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Question 28: moderate

A wooden cube is floating in water with some part inside water. When a stone of mass \(4.5\text{ kg}\) is placed on cube then it further sinks by \(5\text{ cm}\). Then side of cube is:

1. 10 cm
2. 30 cm
3. 60 cm
4. 90 cm
View Answer

The additional weight of the stone is balanced by the extra buoyant force: \(mg = a^2 \Delta x \rho_w g\). Substituting the values: \(4.5 = a^2 (0.05)(1000)\) gives \(a^2 = 0.09\text{ m}^2\), which yields a side length of \(a = 30\text{ cm}\).

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Question 29: moderate

A long capillary is dipped in a beaker containing water. Water rises in capillary upto some height \(h\). Match the statements in list-I with most appropriate effects on water level mentioned in list-II:

**List-I**
(A) Soap solution is added to water
(B) Arrangement taken in a freely falling lift
(C) In a lift accelerating uniformly upward
(D) Arrangement is taken in a lift accelerating uniformly downward

**List-II**
(p) \(h\) decreases
(q) \(h\) increases
(r) \(h\) remains same
(s) water will rise upto complete height of capillary
(t) water level in capillary goes below the outside level

 

1. A - p, B - t, C - q, D - p
2. A - t, B - q, C - s, D - p
3. A - p, B - s, C - p, D - q
4. A - q, B - p, C - s, D - q
View Answer

Soap reduces surface tension, so \(h\) decreases (A-p). In a free fall, effective gravity \(g_{eff} = 0\), so water rises to full height (B-s). Upward acceleration increases \(g_{eff}\) hence \(h\) decreases (C-p). Downward acceleration decreases \(g_{eff}\) hence \(h\) increases (D-q).

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Question 30: moderate

A water tank resting on the floor has two small holes vertically one above the other. The holes are \(h_1\) \(text{cm}\) and \(h_2\) \(text{cm}\) above the floor. How high does water stand in the tank if the jets from the holes hits the floor at the same point ?

1. \(h_1 + h_2\)
2. \(h_2 - h_1\)
3. \(\frac{h_1^2 + h_2^2}{2}\)
4. \(\frac{h_2^2 - h_1^2}{2}\)
View Answer

For equal horizontal range, the height \(H\) of the water level in the tank must satisfy \(h_1(H - h_1) = h_2(H - h_2)\). Solving for \(H\) gives \(H(h_2 - h_1) = h_2^2 - h_1^2\), which simplifies to \(H = h_1 + h_2\).

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