MHT CET 2020 Physics Question Paper with Answer and Solution

(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Since the strings are identical,$\mu$ is the same for both.
For the first string: $N = \frac{1}{2\ell} \sqrt{\frac{T_1}{\mu}}$
For the second string: $1.5N = \frac{1}{2(2\ell)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4\ell} \sqrt{\frac{T_2}{\mu}}$
Dividing the two equations: $\frac{N}{1.5N} = \frac{\frac{1}{2\ell} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4\ell} \sqrt{\frac{T_2}{\mu}}}$
$\frac{1}{1.5} = \frac{4\ell}{2\ell} \sqrt{\frac{T_1}{T_2}}$
$\frac{1}{1.5} = 2 \sqrt{\frac{T_1}{T_2}}$
$\sqrt{\frac{T_1}{T_2}} = \frac{1}{1.5 \times 2} = \frac{1}{3}$
Squaring both sides: $\frac{T_1}{T_2} = \frac{1}{9}$.

(C) The fundamental frequency of a sonometer wire is given by the formula: $n = \frac{1}{2\ell} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$\ell$ is the length,and $m$ is the mass per unit length.
Initially,$n = \frac{1}{2\ell} \sqrt{\frac{T}{m}}$.
When the tension is increased by $8 \ N$,the new tension is $T' = T + 8$. The new frequency is $3n$.
So,$3n = \frac{1}{2\ell} \sqrt{\frac{T+8}{m}}$.
Dividing the second equation by the first equation:
$\frac{3n}{n} = \frac{\frac{1}{2\ell} \sqrt{\frac{T+8}{m}}}{\frac{1}{2\ell} \sqrt{\frac{T}{m}}}$
$3 = \sqrt{\frac{T+8}{T}}$
Squaring both sides:
$9 = \frac{T+8}{T}$
$9T = T + 8$
$8T = 8$
$T = 1 \ N$.

(A) The frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density. Since $L$ and $\mu$ are constant,$n \propto \sqrt{T}$.
In air,the tension $T_1 = mg = V \varrho g$,where $V$ is the volume of the load and $\varrho$ is its specific gravity (density relative to water).
When the load is immersed in water,the buoyant force acts on it. The effective tension becomes $T_2 = V(\varrho - 1)g$ (since the density of water is $1 \text{ g/cm}^3$).
Taking the ratio of the frequencies: $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{V(\varrho - 1)g}{V \varrho g}} = \sqrt{\frac{\varrho - 1}{\varrho}}$.
Therefore,the new frequency $n_2 = n \sqrt{\frac{\varrho - 1}{\varrho}}$.

(B) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since tension $T$ and mass per unit length $\mu$ are constant,$n \propto \frac{1}{L}$.
Let the initial length be $L_1 = L$ and the new length be $L_2 = L + 0.25L = 1.25L$.
The initial frequency $n_1 = 50 \ Hz$.
The new fundamental frequency $n_2$ is given by $\frac{n_2}{n_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.
So,$n_2 = 0.8 \times 50 \ Hz = 40 \ Hz$.
The frequency of the second harmonic is $2n$. Initially,$2n_1 = 2 \times 50 = 100 \ Hz$. Finally,$2n_2 = 2 \times 40 = 80 \ Hz$.
The change in frequency is $100 \ Hz - 80 \ Hz = 20 \ Hz$.
The percentage decrease is $\frac{20}{100} \times 100 \% = 20 \%$.

(D) The fundamental frequency $n$ of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Since $l$ and $\mu$ are constant,$n \propto \sqrt{T}$.
Let the initial tension be $T$ and the final tension be $T + 3$.
Given that the frequency becomes double,$n_2 = 2n_1$.
Using the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{1}{2} = \sqrt{\frac{T}{T+3}}$.
Squaring both sides: $\frac{1}{4} = \frac{T}{T+3}$.
Cross-multiplying: $T + 3 = 4T$.
$3 = 3T$,which gives $T = 1 \ kg$.

(B) For a mass attached to a rigid rod to complete a vertical circle,the minimum velocity at the bottom point is determined by the conservation of mechanical energy.
At the bottom point,the potential energy is $0$ (taking the bottom as the reference level).
At the top point,the height $h = 2L$ and the velocity $v_{top}$ can be $0$ because the rod provides support (unlike a string).
Using the law of conservation of energy:
$E_{bottom} = E_{top}$
$\frac{1}{2} mv^2 = mg(2L) + \frac{1}{2} m(v_{top})^2$
For the minimum velocity,we set $v_{top} = 0$:
$\frac{1}{2} mv^2 = 2mgL$
$v^2 = 4gL$
$v = 2 \sqrt{gL}$

(B) For a conical pendulum (bob whirled in a horizontal plane), the forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$.
Resolving the tension $T$ into vertical and horizontal components:
$1$. The vertical component $T \cos \phi$ balances the weight of the bob: $T \cos \phi = mg$.
$2$. The horizontal component $T \sin \phi$ provides the necessary centripetal force: $T \sin \phi = \frac{mv^{2}}{R}$.
Given:
Mass $m = 100 \,g = 0.1 \,kg$
Tension $T = 10 \,N$
Acceleration due to gravity $g = 10 \,m/s^{2}$
From the vertical equilibrium equation:
$\cos \phi = \frac{mg}{T}$
$\cos \phi = \frac{0.1 \,kg \times 10 \,m/s^{2}}{10 \,N}$
$\cos \phi = \frac{1}{10} = 0.1$
Therefore, the angle $\phi = \cos^{-1}(0.1)$.

(A) The energy stored in a spring stretched by a force $F$ is given by $W = \frac{F^{2}}{2k}$,where $k$ is the spring constant.
For the first spring with constant $K$,the energy stored is $W_{1} = \frac{F^{2}}{2K}$.
For the second spring with constant $2K$,the energy stored is $W_{2} = \frac{F^{2}}{2(2K)} = \frac{F^{2}}{4K}$.
Comparing the two expressions,we see that $W_{1} = \frac{F^{2}}{2K}$ and $W_{2} = \frac{1}{2} \left( \frac{F^{2}}{2K} \right) = \frac{W_{1}}{2}$.
Therefore,$W_{1} = 2W_{2}$.

(C) According to the Work-Energy Theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
$W = \Delta K.E.$
Since the body starts from rest,the initial kinetic energy is $0$.
$W = K.E._{final} - 0 = K.E._{final}$
The work done by a constant force $F$ over a distance $d$ is given by $W = F \cdot d$.
Therefore,$K.E. = F \cdot d$.
Since the force $F$ and the distance $d$ are constants,the kinetic energy acquired is independent of the mass $M$ of the body.

(A) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Given that the kinetic energies of the two masses are equal,we have $K_1 = K_2$.
Substituting the expression for kinetic energy,we get $\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$.
Rearranging the terms to find the ratio of momenta,we get $\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}$.
Taking the square root on both sides,the ratio of momenta is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 1 \ g$ and $m_2 = 4 \ g$,we substitute these values:
$\frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the ratio of their momenta is $1: 2$.

(C) For a body sliding down a smooth inclined plane,the potential energy is converted entirely into translational kinetic energy. Thus,$v = \sqrt{2gh}$.
For a sphere rolling down a rough inclined plane,the potential energy is converted into both translational and rotational kinetic energy. The velocity at the bottom is given by $v_{CM} = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$.
Substituting $v = \sqrt{2gh}$,we get $v_{CM} = \frac{v}{\sqrt{1 + \frac{K^2}{R^2}}}$.
For a solid sphere,the radius of gyration $K$ satisfies $\frac{K^2}{R^2} = \frac{2}{5}$.
Substituting this value,$v_{CM} = \frac{v}{\sqrt{1 + \frac{2}{5}}} = \frac{v}{\sqrt{\frac{7}{5}}} = \sqrt{\frac{5}{7}} v$.

(D) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation:
$K = \frac{p^2}{2m}$
Rearranging this for momentum,we get:
$p = \sqrt{2mK}$
Since the kinetic energy $K$ is the same for both bodies,the momentum $p$ is directly proportional to the square root of the mass:
$p \propto \sqrt{m}$
Because the heavy body has a larger mass $m$ compared to the light body,it will have a greater momentum $p$.

(C) The work done by a force is defined as $W = \vec{F} \cdot \vec{s} = Fs \cos \theta$,where $\theta$ is the angle between the force and the displacement.
In this case,the normal reaction force $N$ acts vertically upwards,perpendicular to the horizontal surface.
The displacement $s$ of the block is along the horizontal surface.
Therefore,the angle $\theta$ between the normal reaction $N$ and the displacement $s$ is $90^\circ$.
Since $\cos 90^\circ = 0$,the work done by the normal reaction is $W = Ns \cos 90^\circ = 0$.

(C) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d}$.
First,calculate the displacement vector $\vec{d} = \vec{r_2} - \vec{r_1}$.
$\vec{d} = (6 - 3) \hat{\imath} + (-1 - 2) \hat{\jmath} + (4 - (-1)) \hat{k} = 3 \hat{\imath} - 3 \hat{\jmath} + 5 \hat{k} \text{ m}$.
Now,calculate the work done:
$W = (5 \hat{\imath} - 2 \hat{\jmath} + 3 \hat{k}) \cdot (3 \hat{\imath} - 3 \hat{\jmath} + 5 \hat{k})$.
$W = (5 \times 3) + (-2 \times -3) + (3 \times 5) = 15 + 6 + 15 = 36 \text{ J}$.