KVPY 2017 Physics Question Paper with Answer and Solution

(D) The force exerted by the water on the marble is due to the hydrostatic pressure at the depth of the marble. The pressure $P$ at the bottom of the bucket is given by $P = \rho g h$,where $\rho = 1000 \, kg/m^3$ is the density of water,$g = 10 \, m/s^2$ is the acceleration due to gravity,and $h = 10 \, cm = 0.1 \, m$ is the height of the water column.
The force $F$ acting downwards on the marble due to the water pressure is $F = P \times A$,where $A$ is the cross-sectional area of the hole (which is equal to the cross-sectional area of the marble at its equator).
The area $A = \pi r^2$,where $r = 1 \, cm = 0.01 \, m$.
Substituting the values:
$F = (1000 \, kg/m^3) \times (10 \, m/s^2) \times (0.1 \, m) \times \pi \times (0.01 \, m)^2$
$F = 1000 \times 10 \times 0.1 \times 3.14159 \times 0.0001$
$F = 1000 \times 0.000314159 \approx 0.314 \, N$.
Since the pressure acts downwards on the top surface of the marble,the net force due to water is $0.31 \, N$ downwards.

(A) The given equation is $\frac{V}{t} = k \left( \frac{p}{l} \right)^a \eta^b r^c$.
The dimensions of the quantities are:
$[V/t] = [L^3 T^{-1}]$
$[p/l] = [M L^{-1} T^{-2} / L] = [M L^{-2} T^{-2}]$
$[\eta] = [M L^{-1} T^{-1}]$
$[r] = [L]$
Substituting these into the equation:
$[L^3 T^{-1}] = [M L^{-2} T^{-2}]^a [M L^{-1} T^{-1}]^b [L]^c$
$[L^3 T^{-1}] = M^{a+b} L^{-2a-b+c} T^{-2a-b}$
Equating the powers of $M, L,$ and $T$ on both sides:
For $M$: $a + b = 0 \Rightarrow b = -a$
For $T$: $-2a - b = -1$
Substituting $b = -a$ into the $T$ equation: $-2a - (-a) = -1 \Rightarrow -a = -1 \Rightarrow a = 1$.
Since $b = -a$,we get $b = -1$.
For $L$: $-2a - b + c = 3$
Substituting $a = 1$ and $b = -1$: $-2(1) - (-1) + c = 3 \Rightarrow -2 + 1 + c = 3 \Rightarrow -1 + c = 3 \Rightarrow c = 4$.
Thus,the values are $a=1, b=-1, c=4$.