TS EAMCET 2020 Physics Question Paper with Answer and Solution

(A) The relationship between conservative force $F$ and potential energy $U(x)$ is given by $F = -\frac{dU}{dx}$.
This means the force is equal to the negative of the slope of the $U(x)$ versus $x$ graph.
For the region $OA$ (from $x=0$ to $x=P$),the graph is a straight line with a positive constant slope. Let the slope be $k > 0$. Thus,$F = -k$,which is a negative constant.
For the region $AB$ (for $x > P$),the graph is a horizontal straight line. The slope of this line is $0$. Thus,$F = -(0) = 0$.
Comparing this with the given options,the graph that shows a negative constant force for $x < P$ and zero force for $x > P$ is represented by option $A$.

(B) Let the mass of each block be $m$. The system moves with a common acceleration $a = \frac{F}{2m}$.
In the frame of the centre of mass,the blocks are at rest at the moment of maximum extension. The pseudo force acting on each mass is $F_p = ma = \frac{F}{2}$.
Let $x_1$ and $x_2$ be the displacements of the two blocks from their initial positions in the centre of mass frame. The total extension of the spring is $x = x_1 + x_2$.
The work done by the external force and pseudo forces in the centre of mass frame is equal to the change in potential energy of the spring:
$W = \frac{1}{2} k x^2$
In the centre of mass frame,the work done by the effective forces is:
$(F - ma)x_1 + (ma)x_2 = \frac{1}{2} k (x_1 + x_2)^2$
Substituting $a = \frac{F}{2m}$:
$(F - \frac{F}{2})x_1 + (\frac{F}{2})x_2 = \frac{1}{2} k (x_1 + x_2)^2$
$\frac{F}{2}(x_1 + x_2) = \frac{1}{2} k (x_1 + x_2)^2$
$x_1 + x_2 = \frac{F}{k}$
Given $F = 10 \,N$ and $k = 2500 \,N/m$,the maximum extension is:
$x_{max} = \frac{10}{2500} \,m = 0.004 \,m = 0.4 \,cm$
The maximum distance between the blocks is the sum of the natural length and the maximum extension:
$d_{max} = 10 \,cm + 0.4 \,cm = 10.4 \,cm$.

(B) The law of conservation of energy is valid in both macroscopic and microscopic domains. Therefore, the statement that it is valid only in the macroscopic domain is incorrect.
All conserved quantities are not necessarily scalars. For example, the conservation of linear momentum and angular momentum involve vector quantities, whereas the conservation of energy involves a scalar quantity.

(D) The block is moving on a rough horizontal surface. The forces acting on the block are the normal force $N$,gravitational force $mg$,the pulling force $F$,and the kinetic frictional force $fr_k$.
For vertical equilibrium,the normal force is $N = mg$.
The kinetic frictional force is given by $fr_k = \mu N = \mu mg$.
Power $P$ is defined as the product of the applied force $F$ and the velocity $v$ in the direction of motion: $P = F v$.
For the block to move at a constant velocity (which is its maximum velocity),the net force on the block must be zero.
Therefore,the pulling force $F$ must be equal to the kinetic frictional force $fr_k$:
$F = fr_k = \mu mg$.
Substituting this into the power equation:
$P = F v_{\max} = (\mu mg) v_{\max}$.
Solving for $v_{\max}$:
$v_{\max} = \frac{P}{\mu mg}$.

(D) Given,mass $m = 10 \ kg$ and displacement $x = \frac{t^3}{25} \ m$.
Velocity $v = \frac{dx}{dt} = \frac{3t^2}{25} \ m/s$.
Acceleration $a = \frac{dv}{dt} = \frac{6t}{25} \ m/s^2$.
Force $F = m \cdot a = 10 \cdot \left(\frac{6t}{25}\right) = \frac{12t}{5} \ N$.
Work done $dW = F \cdot dx = F \cdot \left(\frac{dx}{dt}\right) dt = \left(\frac{12t}{5}\right) \cdot \left(\frac{3t^2}{25}\right) dt = \frac{36t^3}{125} dt$.
Total work done in the first $2 \ s$ is $W = \int_{0}^{2} \frac{36t^3}{125} dt = \frac{36}{125} \left[\frac{t^4}{4}\right]_{0}^{2} = \frac{36}{125} \cdot \frac{16}{4} = \frac{36 \cdot 4}{125} = \frac{144}{125} = 1.152 \ J$.

(C) According to the Work-Energy Theorem,the work done by the net (resultant) force acting on a body is equal to the change in its kinetic energy.
Mathematically,$W_{net} = \Delta KE = KE_f - KE_i$.
Therefore,the correct option is $C$.

(C) The displacement of the object is along the $X$-axis by $3 \text{ m}$, so the displacement vector is $\vec{d} = 3 \hat{i} \text{ m}$.
Given the force vector is $\vec{F} = (2 \hat{i} + 4 \hat{j}) \text{ N}$.
The work done $W$ by a constant force is given by the dot product of the force vector and the displacement vector:
$W = \vec{F} \cdot \vec{d}$
$W = (2 \hat{i} + 4 \hat{j}) \cdot (3 \hat{i})$
$W = (2 \times 3)(\hat{i} \cdot \hat{i}) + (4 \times 0)(\hat{j} \cdot \hat{i})$
Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = 0$, we get:
$W = 6 \times 1 + 0 = 6 \text{ J}$.

(C) Mass of the elevator,$m = 500 \,kg$.
Acceleration of the elevator,$a = 2 \,m/s^2$ (upward).
When the elevator is moving upward,the tension $T$ in the cable is given by the equation of motion: $T - mg = ma$.
Therefore,$T = m(g + a) = 500(10 + 2) = 500(12) = 6000 \,N$.
The work done by the tension force $W$ is given by $W = T \times s$,where $s = 12 \,m$ is the displacement.
$W = 6000 \,N \times 12 \,m = 72000 \,J$.
Converting to kilojoules,$W = 72 \,kJ$.

(D) Mass of the body, $m = 10 \,kg$. Force, $F = 4 \,N$. Acceleration produced in the body due to the applied force is $a = \frac{F}{m} = \frac{4}{10} = 0.4 \,m/s^2$.
For the interval $0 \leq t \leq 1 \,s$, the distance traveled $s_1$ is $s_1 = u t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 0.4 \times (1)^2 = 0.2 \,m$.
Work done $W_1 = F \times s_1 = 4 \times 0.2 = 0.8 \,J$.
For the interval $1 \,s \leq t \leq 2 \,s$, the initial velocity $v_i$ at $t = 1 \,s$ is $v_i = u + a t = 0 + 0.4 \times 1 = 0.4 \,m/s$.
The distance traveled $s_2$ during this interval $(t = 1 \,s)$ is $s_2 = v_i t + \frac{1}{2} a t^2 = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times (1)^2 = 0.4 + 0.2 = 0.6 \,m$.
Work done $W_2 = F \times s_2 = 4 \times 0.6 = 2.4 \,J$.
The ratio $\frac{W_2}{W_1} = \frac{2.4}{0.8} = 3$.

(A) Given: $m = 2 \,kg$, $x = \frac{\alpha t^2}{2}$, and $\alpha = 1 \,m/s^2$.
First, we find the velocity $v$ by differentiating the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha t^2}{2} \right) = \alpha t$.
At $t = 2 \,s$, the velocity $v$ is:
$v = 1 \times 2 = 2 \,m/s$.
According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
At $t = 0$, $v_i = \alpha(0) = 0$.
At $t = 2 \,s$, $v_f = 2 \,m/s$.
Therefore, $W = \frac{1}{2} \times 2 \,kg \times (2 \,m/s)^2 - 0 = 4 \,J$.

(A) Given,applied force,$F=10 \,N$.
Mass of the block,$m=2 \,kg$.
Coefficient of kinetic friction,$\mu=0.2$.
Normal force,$N=mg=2 \times 10=20 \,N$.
Friction force,$f=\mu N=0.2 \times 20=4 \,N$.
Net force,$F_{\text{net}}=F-f=10-4=6 \,N$.
Acceleration,$a=\frac{F_{\text{net}}}{m}=\frac{6}{2}=3 \,m/s^2$.
Distance covered in $t=4 \,s$ starting from rest $(u=0)$:
$s=ut+\frac{1}{2}at^2=0+\frac{1}{2} \times 3 \times (4)^2=24 \,m$.
Work done by the applied force of $10 \,N$ $(W_1)$:
$W_1=F \times s=10 \times 24=240 \,J$.
Work done by the friction force of $4 \,N$ $(W_2)$:
$W_2=f \times s \times \cos(180^{\circ})=-4 \times 24=-96 \,J$.

(A) Given,mass of particle,$m = 30 \,g = 3 \times 10^{-2} \,kg$.
The position of the particle is $x = \alpha t^2$.
Velocity $v = \frac{dx}{dt} = 2\alpha t$.
Acceleration $a = \frac{dv}{dt} = 2\alpha$.
Force $F = ma = (3 \times 10^{-2} \,kg) \times (2 \times 1 \,m/s^2) = 6 \times 10^{-2} \,N$.
Work done $W = \int F dx$. Since $x = \alpha t^2$,$dx = 2\alpha t dt$.
At $t = 0, x = 0$. At $t = 4, x = 1 \times (4)^2 = 16 \,m$.
$W = \int_{0}^{16} (6 \times 10^{-2}) dx = 6 \times 10^{-2} \times [x]_{0}^{16} = 6 \times 10^{-2} \times 16 = 0.96 \,J$.

(B) The colours observed in a soap bubble are produced due to the phenomenon of interference.
When light falls on the thin film of the soap bubble,it reflects from both the outer and inner surfaces.
These reflected waves undergo superposition,leading to constructive or destructive interference depending on the thickness of the film and the wavelength of the light.
This interference pattern results in the appearance of various colours.

(A) point source emits light waves in all possible directions in a three-dimensional space. Since the speed of light is constant in a homogeneous medium,the locus of all points vibrating in the same phase at a given time forms a sphere centered at the source. Therefore,the wavefront originating from a point source is spherical.

(C) In Young's double slit experiment,the fringe width is given by $\beta = \frac{\lambda D}{d} \Rightarrow \beta \propto \frac{1}{d}$. So,if the distance between the two slits $(d)$ increases,the fringe width reduces. This statement is correct.
$(b)$ In diffraction due to a single slit,the path difference between rays reaching the center of the screen is zero,so its intensity at the center is maximum,i.e.,a bright fringe is observed at the center. This statement is correct.
$(c)$ The resolving power of a microscope is defined as $\frac{1}{d_{\min }} = \frac{2 n \sin \theta}{1.22 \lambda}$. It is the reciprocal of the minimum separation $(d_{\min })$ of two points seen as distinct,not the maximum separation. Thus,this statement is incorrect.
$(d)$ Polarisation is possible only in those waves that have vibrations in different directions,which is the case for transverse waves. This statement is correct.
Therefore,the incorrect statement is $(c)$.

(D) In Young's double slit experiment $(YDSE)$,the fringe width $\beta$ is given by the formula: $\beta = \frac{\lambda D}{d}$.
Here,$\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
Since $D$ and $d$ are constant,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.
The wavelengths of the given colors follow the order: $\lambda_R > \lambda_G > \lambda_B$.
Therefore,the fringe widths will follow the same order: $\beta_R > \beta_G > \beta_B$.

(D) Given: $\lambda_1 = 6600 \ \text{Å}$,$\lambda_2 = 4400 \ \text{Å}$,$n_1 = 60$.
In Young's double slit experiment,the angular width of the field of view is constant.
The number of fringes $n$ is inversely proportional to the fringe width $\beta$,where $\beta = \frac{\lambda D}{d}$.
Thus,$n \propto \frac{1}{\lambda}$,which implies $n_1 \lambda_1 = n_2 \lambda_2$.
Substituting the given values:
$60 \times 6600 = n_2 \times 4400$
$n_2 = \frac{60 \times 6600}{4400}$
$n_2 = 60 \times \frac{66}{44} = 60 \times 1.5 = 90$.
Therefore,$90$ fringes will be observed.

(C) The fringe width $\beta$ in Young's double slit experiment is given by the formula:
$\beta = \frac{D \lambda}{d}$
where $D$ is the distance between the slits and the screen,$\lambda$ is the wavelength of light,and $d$ is the slit separation.
Given that the new slit separation $d_2 = 2d_1$ and the new distance $D_2 = \frac{D_1}{2}$.
The new fringe width $\beta_2$ is:
$\beta_2 = \frac{D_2 \lambda}{d_2} = \frac{(D_1 / 2) \lambda}{2 d_1} = \frac{1}{4} \left( \frac{D_1 \lambda}{d_1} \right)$
$\beta_2 = \frac{1}{4} \beta_1$
Therefore,the fringe width becomes $\frac{1}{4}$ times its original value.

(C) Given: Velocity of the conducting bar $v=5 \ m \ s^{-1}$,Magnetic field $B=0.1 \ T$.
At time $t=4 \ s$,the distance of the bar from the vertex is $x = v \times t = 5 \times 4 = 20 \ m$.
Since the rails form a right angle $(90^{\circ})$,the triangle formed by the rails and the bar is an isosceles right-angled triangle.
The length of the bar $l$ in contact with the rails is $l = 2 \times x \tan(45^{\circ}) = 2 \times 20 \times 1 = 40 \ m$.
The induced emf is given by $e = B \times v \times l$.
Substituting the values: $e = 0.1 \times 5 \times 40 = 20 \ V$.

(D) For the first lens:
$f_1 = +10 \ cm, u_1 = -30 \ cm$.
Applying the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.
So,$v_1 = +15 \ cm$.
This image acts as an object for the second lens. The distance between the first and second lens is $5 \ cm$,so the object distance for the second lens is $u_2 = +(15 - 5) = +10 \ cm$.
For the second lens $(f_2 = -10 \ cm)$:
$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-10} + \frac{1}{10} = 0$.
So,$v_2 = \infty$.
The rays emerging from the second lens are parallel to the principal axis.
These parallel rays fall on the third lens $(f_3 = +30 \ cm)$. Parallel rays incident on a convex lens converge at its focus.
Therefore,the final image is formed at a distance of $30 \ cm$ from the third lens.