Q.50 Projectiles A and B are thrown from the top of a 400 m high tower at angles
45^{\circ} and 60^{\circ} with the vertical respectively.
If their ranges and times of flight are the same, the ratio of their speeds of projection
v_{A}:v_{B} is:
(1) 1:\sqrt{3}
(2) \sqrt{2}:1
(3) 1:2
(4) 1:\sqrt{2}
Answer:
Bonus (No correct option)
Solution:
Both projectiles are thrown from the same height.
Time of flight depends on the vertical component of velocity and the height of projection.
Range depends on both horizontal component of velocity and time of flight.
For projectile A:
Vertical component =v_{A}\cos45^{\circ}
Horizontal component =v_{A}\sin45^{\circ}
For projectile B:
Vertical component =v_{B}\cos60^{\circ}
Horizontal component =v_{B}\sin60^{\circ}
If time of flight is the same, the vertical components must be the same.
If range is also the same, the horizontal components must be the same.
These two conditions cannot be satisfied simultaneously for angles
45^{\circ} and 60^{\circ}.
Hence, no value of v_{A}:v_{B} satisfies both conditions together.
Therefore, all given options are incorrect.
Q.51 A power transmission line feeds input power at 2.3 kV to a step-down transformer with its primary winding having 3000 turns.
The output power is delivered at 230 V by the transformer.
The current in the primary of the transformer is 5 A and its efficiency is 90\%.
The winding of the transformer is made of copper.
The output current of the transformer is ____ A.
Answer:
45
Solution:
Input power to the transformer is:
P_i = 2300 \times 5Output power considering efficiency is:
P_o = 2300 \times 5 \times 0.9Output power is also given by:
P_o = 230 \times I_2Equating:
230 \times I_2 = 2300 \times 5 \times 0.9Solving:
I_2 = 45Hence, the output current of the transformer is 45 A.
Q.52 A big drop is formed by coalescing 1000 small identical drops of water.
If E_{1} is the total surface energy of the 1000 small drops and
E_{2} is the surface energy of the single big drop, then
The value of x is ____.
Answer:
10
Solution:
Volume is conserved during coalescence.
1000 \times \frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3} R^{3}=1000r^{3} R=10rSurface energy is proportional to surface area.
Total surface energy of 1000 small drops is:
E_{1}=1000 \times 4\pi r^{2} \times SSurface energy of the big drop is:
E_{2}=4\pi(10r)^{2}\times SRatio of surface energies is:
\frac{E_{1}}{E_{2}}=\frac{1000 \times 4\pi r^{2}}{4\pi(100r^{2})} \frac{E_{1}}{E_{2}}=\frac{10}{1}Hence,
x=10Q.53 Two discs have moments of inertia
I_{1}=4\ \text{kg m}^{2} and I_{2}=2\ \text{kg m}^{2}
about their central axes and normal to their planes.
They are rotating with angular speeds
10\ \text{rad s}^{-1} and 4\ \text{rad s}^{-1} respectively.
The discs are brought into contact face to face with their axes of rotation coincident.
The loss in kinetic energy of the system in the process is ____ J.
Answer:
24
Solution:
Since there is no external torque, angular momentum is conserved.
I_{1}\omega_{1}+I_{2}\omega_{2}=(I_{1}+I_{2})\omega_{0}Substituting values:
4\times10+2\times4=6\omega_{0} 48=6\omega_{0} \omega_{0}=8\ \text{rad s}^{-1}Initial kinetic energy of the system is:
E_{1}=\frac{1}{2}I_{1}\omega_{1}^{2}+\frac{1}{2}I_{2}\omega_{2}^{2} E_{1}=\frac{1}{2}\times4\times10^{2}+\frac{1}{2}\times2\times4^{2} E_{1}=200+16 E_{1}=216\ \text{J}Final kinetic energy after contact is:
E_{2}=\frac{1}{2}(I_{1}+I_{2})\omega_{0}^{2} E_{2}=\frac{1}{2}\times6\times8^{2} E_{2}=192\ \text{J}Loss in kinetic energy is:
\Delta E=E_{1}-E_{2} \Delta E=216-192 \Delta E=24\ \text{J}Hence, the loss in kinetic energy is 24 J.
Q.54 In an experiment to measure the focal length f of a convex lens, the magnitudes of object distance x and image distance y are measured with reference to the focal point of the lens.
The y-x plot is shown in the figure.
The focal length of the lens is ____ cm.
Answer:
20
Solution:
From the graph, the intercepts on the x-axis and y-axis are both equal to 20 cm.
Using the lens formula with distances measured from the focal point:
\frac{1}{f+20}-\frac{1}{-(f+20)}=\frac{1}{f}Simplifying:
\frac{2}{f+20}=\frac{1}{f}Solving:
2f=f+20 f=20\ \text{cm}Alternatively, from the y-x graph relation:
x_{1}x_{2}=f^{2} 20\times20=f^{2} f=20\ \text{cm}Hence, the focal length of the lens is 20 cm.
Q.55 A vector has magnitude equal to that of
\vec{A}=3\hat{i}+4\hat{j}
and is parallel to
\vec{B}=4\hat{i}+3\hat{j}.
The x and y components of this vector in the first quadrant are x and 3 respectively, where
x= ____.
Answer:
4
Solution:
Magnitude of vector \vec{A} is:
|\vec{A}|=\sqrt{3^{2}+4^{2}} |\vec{A}|=5Unit vector along \vec{B} is:
\hat{B}=\frac{4\hat{i}+3\hat{j}}{\sqrt{4^{2}+3^{2}}} \hat{B}=\frac{4\hat{i}+3\hat{j}}{5}Required vector is:
\vec{N}=|\vec{A}|\hat{B} \vec{N}=5\times\frac{4\hat{i}+3\hat{j}}{5} \vec{N}=4\hat{i}+3\hat{j}Hence, the x-component is:
x=4