Question 41:
moderateA 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 :
The horizontal range of water emerging from a hole is given by the formula
$$R = 2\sqrt{h(H - h)} $$
Question 42:
easyTwo 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  Â
\(r^2 \), 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 \( \frac{v_B}{v_A} \)Â is 4 (Option 3).
Question 43:
easyThe 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, \(u = \frac{1}{2} \times (2.0 \times 10^{11}) \times (10^{-3})^2 = 10^5\text{ J/m}^3\).
Question 44:
easyThe 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 \( F = 6 \times 3.14 \times 0.8 \times 0.5 \times 10^{-3} \times 2 \approx 15 \times 10^{-3}\text{ N}\).
Question 45:
moderateA 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
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 \(T' = T\sqrt{\frac{g}{g'}} = \sqrt{3}\sqrt{\frac{4}{3}} = 2\text{ s}\).
Question 46:
moderateA 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:
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 \(\Delta L = \frac{\rho g L^2}{2Y}\).
Question 47:
moderateA 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:
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 \(\frac{3\%}{3} = 1\%\).
Question 48:
moderateA 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:
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 \(a = 30\text{ cm}\).
Question 49:
easyIf 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 \(F = T \times 4\pi R\).
Question 50:
easyA 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 \(v' = 3^2 v_0 = 9v_0\).