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:
1. \(\left(\frac{\sigma}{\rho} - 1\right)h\)
2. \(\left(\frac{\sigma}{\rho} + 1\right)h\)
3. \(\left(\frac{\rho}{\sigma} - 1\right)h\)
4. \(\left(\frac{\rho}{\sigma} + 1\right)h\)
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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 \(H = \left(\frac{\sigma}{\rho} - 1\right)h\).
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. \(\frac{h_2^2 - h_1^2}{2}\)
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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 \(H = h_1 + h_2\).
A liquid is flowing in a cylindrical pipe of internal diameter \(4\text{ cm}\) with a velocity of \(5\text{ m/s}\). If this tube is joined with another tube of internal diameter \(2\text{ cm}\) then the velocity of flow of liquid in the smaller tube will be (in \(\text{ms}^{-1}\))
View Answer
Using the equation of continuity, \(A_1 v_1 = A_2 v_2\), which ⇒ \(d_1^2 v_1 = d_2^2 v_2\). Substituting \(d_1 = 4\text{ cm}\), \(v_1 = 5\text{ m/s}\), and \(d_2 = 2\text{ cm}\) gives \(16 \times 5 = 4 \times v_2\), so \(v_2 = 20\text{ m/s}\).
Terminal velocity of iron ball of radius \(1\text{ mm}\) in glycerine is \(v_1\) and \(v_2\) is the terminal velocity of lead ball of radius \(2\text{ mm}\) in glycerine then \(v_1 : v_2\) is : [\(\rho_{\text{glycerine}} = 13.6\text{ g/cc}\), \(\sigma_{\text{Fe}} = 7.6\text{ g/cc}\), \(\sigma_{\text{Pb}} = 11.6\text{ g/cc}\)]
1. 0.47
2. 0.58
3. 0.75
4. 0.21
View Answer
Terminal velocity \(v_T \propto r^2(\sigma - \rho)\). For iron, \(v_1 \propto 1^2(7.6 - 13.6) = -6\). For lead, \(v_2 \propto 2^2(11.6 - 13.6) = -8\). Thus, the ratio is \(v_1 : v_2 = -6 / -8 = 0.75\).
A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is:
1. P
2. \(\frac{3P}{4}\)
3. \(\frac{P}{2}\)
4. \(\frac{P}{4}\)
View Answer
Rate of flow \(Q = \frac{\pi P r^4}{8 \eta l}\). Under new conditions, \(Q' = 2Q\), \(r' = 2r\), and \(l' = 2l\), giving \(Q' = \frac{\pi P' (2r)^4}{8 \eta (2l)} = 8 \left(\frac{\pi P' r^4}{8 \eta l}\). Thus, \(8 P' = 2 P\), yielding \(P' = P/4\).
Consider the following statements:
Statement A: When a capillary tube is dipped into a liquid, if the liquid neither rises nor falls in the capillary, then the angle of contact must be zero.
Statement B: The working of a Venturi-meter is based on Bernoulli’s theorem.
Based upon the above information, pick the correct option:
1. Both statement (A) and (B) are correct
2. Both statement (A) and (B) are incorrect
3. Statement (A) is correct while statement (B) is incorrect
4. Statement (A) is incorrect while statement (B) is correct
View Answer
If liquid neither rises nor falls, the net vertical surface tension force is zero, meaning the contact angle is \(90^\circ\). Venturi-meter works on Bernoulli's theorem, so only statement B is correct.
A small ball of mass \(m\) and density \(rho\) is dropped in a viscous liquid of density \(\rho_0\). After sometime, the ball falls with constant velocity. The net force acting on the ball after it attains constant velocity, will be:
1. \(mg\left(\frac{\rho_0}{\rho}-1\right)\)
2. \(mg(\rho - \rho_0)\)
3. \(mg\left(1-\frac{\rho}{\rho_0}\right)\)
4. Zero
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When a body falls with constant velocity (terminal velocity), its acceleration is zero. Thus, by Newton's second law, the net force on it is zero.
Assertion (A): Bernoulli’s theorem is based on energy conservation.
Reason (R): Bernoulli’s theorem holds good for all liquids.
1. Both Assertion & Reason are true and the reason is the correct explanation of the assertion.
2. Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.
3. Assertion is true statement but Reason is false.
4. Both Assertion and Reason are false statements.
View Answer
Bernoulli's theorem is based on energy conservation (Assertion is true). However, it only holds for ideal (non-viscous, incompressible) fluids, not all liquids (Reason is false).
Assertion (A): Bernoulli’s theorem is based on energy conservation.
Reason (R): Bernoulli’s theorem holds good for all liquids.
1. Both Assertion & Reason are true and the reason is the correct explanation of the assertion.
2. Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.
3. Assertion is true statement but Reason is false.
4. Both Assertion and Reason are false statements.
View Answer
Bernoulli's theorem is derived from the law of conservation of energy (Assertion is true). It only holds good for ideal (non-viscous, incompressible) fluids, not all liquids (Reason is false).
Assertion (A): Blood pressure of heart is same whether you lie down or stand up.
Reason (R): Pressure varies with height in a fluid under gravity.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true due to physiological regulation of blood pressure at the heart level. Reason (R) is true as pressure in a fluid varies with depth, \(P = \rho gh\). However, (R) describes a physical effect that (A) counteracts, so (R) is not the correct explanation for (A).