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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