In a shotput event an athlete throws
Question: In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1 m/s at 45ofrom height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m/s2, the kinetic energy of the shotput when it just reaches the ground will be (a) 2.5 J (b) 5.0 J (c) 52.5 J (d) 155.0 J Solution: (d) 155.0 J...
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Question: If $\cos \mathrm{X}=\frac{-3}{5}$ and $\pi\mathrm{x}\frac{3 \pi}{2}$ find the values of tan 2x Solution: To find: tan2x From part (i) and (ii), we have $\sin 2 x=\frac{24}{25}$ and $\cos 2 \mathrm{x}=-\frac{7}{25}$ We know that, $\tan x=\frac{\sin x}{\cos x}$ Replacing x by 2x, we get $\tan 2 x=\frac{\sin 2 x}{\cos 2 x}$ Putting the values of sin 2x and cos 2x, we get $\tan 2 x=\frac{\frac{24}{25}}{-\frac{7}{25}}$ $\tan 2 x=\frac{24}{25} \times\left(-\frac{25}{7}\right)$ $\therefore \t...
Read More →A body of mass 0.5 kg travels
Question: A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2where a = 5 m-1/2s-1. The work done by the net force during its displacement from x = 0 to x = 2 m is (a) 1.5 J (b) 50 J (c) 10 J (d) 100 J Solution: (b) 50 J...
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Question: If $\cos \mathrm{x}=\frac{-3}{5}$ and $\pi\mathrm{x}\frac{3 \pi}{2}$ find the values of cos 2x Solution: Given: $\cos \mathrm{x}=\frac{-3}{5}$ To find: $\cos 2 x$ We know that $\cos 2 x=2 \cos ^{2} x-1$ Putting the value, we get $\cos 2 x=2\left(-\frac{3}{5}\right)^{2}-1$ $\cos 2 x=2 \times \frac{9}{25}-1$ $\cos 2 x=\frac{18}{25}-1$ $\cos 2 x=\frac{18-25}{25}$ $\therefore \cos 2 \mathrm{x}=-\frac{7}{25}$...
Read More →During inelastic collision between two bodies,
Question: During inelastic collision between two bodies, which of the following quantities always remain conserved? (a) total kinetic energy (b) total mechanical energy (c) total linear momentum (d) speed of each body Solution: (c) total linear momentum...
Read More →A body is falling freely under the action of gravity
Question: A body is falling freely under the action of gravity alone in a vacuum. Which of the following quantities remain constant during the fall? (a) kinetic energy (b) potential energy (c) total mechanical energy (d) total linear momentum Solution: (c) total mechanical energy...
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Question: If $\cos \mathrm{x}=\frac{-3}{5}$ and $\pi\mathrm{x}\frac{3 \pi}{2}$ find the values of sin 2x Solution: Given: $\cos x=\frac{-3}{5}$ To find: $\sin 2 x$ We know that, sin2x = 2 sinx cosx (i) Here, we dont have the value of sin x. So, firstly we have to find the value of sinx We know that $\cos ^{2} x+\sin ^{2} x=1$ Putting the values, we get $\left(-\frac{3}{5}\right)^{2}+\sin ^{2} x=1$ $\Rightarrow \frac{9}{25}+\sin ^{2} x=1$ $\Rightarrow \sin ^{2} x=1-\frac{9}{25}$ $\Rightarrow \sin...
Read More →A bicyclist comes to a skidding stop in 10 m.
Question: A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is (a) +2000 J (b) -200 J (c) zero (d) -20,000 J Solution: (c) As the friction is present in fhis problem, so mechanical energy is not conserved. So energy will be lost due to dissipation by friction.^Here, work is done by the frictional force on the cycle and is equal to __ 200 x 10 = -2000 J ...
Read More →A man squatting on the ground gets straight up and stand.
Question: A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is (a) constant and equal to mg in magnitude (b) constant and greater than mg in magnitude (c) variable but always greater than mg (d) at first greater than mg and later becomes equal to mg Solution: (d) In the process of squatting on the ground he gets straight up and stand. Then he is tilted somewhat, the man exerts a variable force on the ground to balance his weig...
Read More →A proton is kept at rest.
Question: A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is (a) same as the same force law is involved in the two experiments (b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens (c) more for the case of a positr...
Read More →An electron and a proton are moving
Question: An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because, (a) the two magnetic forces are equal and opposite, so they produce no net effect (b) the magnetic forces do no work on each particle (c) the magnetic forces do equal and opposite work on each particle (d) the magnetic forces are necessarily negligible Solution: (b) the m...
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Question: If $\sin \mathrm{x}=\frac{\sqrt{5}}{3}$ and $0\mathrm{x}\frac{\pi}{2}$ find the values of $\tan 2 x$ Solution: To find: tan2x From part (i) and (ii), we have $\sin 2 x=\frac{4 \sqrt{5}}{9}$ And $\cos 2 \mathrm{x}=-\frac{1}{9}$ We know that, $\tan x=\frac{\sin x}{\cos x}$ Replacing x by 2x, we get $\tan 2 x=\frac{\sin 2 x}{\cos 2 x}$ Putting the values of sin 2x and cos 2x, we get $\tan 2 x=\frac{\frac{4 \sqrt{5}}{9}}{-\frac{1}{9}}$ $\tan 2 \mathrm{x}=\frac{4 \sqrt{5}}{9} \times(-9)$ $\...
Read More →A helicopter of mass 2000 kg rises with a vertical
Question: A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s2. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the (a) force on the floor of the helicopter by the crew and passengers (b) action of the rotor of the helicopter on the surrounding air (c) force on the helicopter due to the surrounding air Solution: Given, Mass of helicopter, M = 2000 kg Mass of the crew and passenger, m = 500 kg Acceleration of helicopter with crew and ...
Read More →A cricket bowler releases the ball in two different ways
Question: A cricket bowler releases the ball in two different ways (a) giving it only horizontal velocity and (b) giving it horizontal velocity and a small downward velocity. The speed vsat the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance. Solution: (a) $\frac{1}{2} v_{z}^{2}=g H \Rightarrow v_{z}=\sqrt{2 g H}$ Speed at ground is given as: $\sqrt{v_{s}^{2}+v_{z}^{2}}=\sqrt{v_{...
Read More →There are three forces F1, F2, and F3 acting on a body,
Question: There are three forces F1, F2, and F3acting on a body, all acting on a point P on the body. The body is found to move with uniform speed. (a) show that the forces are coplanar (b) show that the torque acting on the body about any point due to these three forces is zero Solution: (a) Acceleration of the body is zero as the three forces F1, F2, and F3on a point on a body has a resultant force equal to zero. The direction of forces F1and F2are in the plane of the paper and in their sum is...
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Question: If $\sin \mathrm{x}=\frac{\sqrt{5}}{3}$ and $0\mathrm{x}\frac{\pi}{2}$ find the values of $\cos 2 x$ Solution: Given: $\sin \mathrm{x}=\frac{\sqrt{5}}{3}$ To find: $\cos 2 x$ We know that $\cos 2 x=1-2 \sin ^{2} x$ Putting the value, we get $\cos 2 x=1-2\left(\frac{\sqrt{5}}{3}\right)^{2}$ $\cos 2 x=1-2 \times \frac{5}{9}$ $\cos 2 x=1-\frac{10}{9}$ $\cos 2 x=\frac{9-10}{9}$ $\therefore \cos 2 x=-\frac{1}{9}$...
Read More →A person in an elevator accelerating upwards
Question: A person in an elevator accelerating upwardswith an acceleration of 2 m/s2, tosses a coin vertically upwards with a speed of 20 m/s. After how much time will the coin fall back into his hand? Solution: The upward acceleration of an elevator, a = 2 m/s2 Acceleration due to gravity, g = 10 m/s2 Therefore, the net effective acceleration, a = (a + g) = 12 m/s2 Considering the effective motion of the coin, v = 0 t = time taken by the coin to achieve maximum height u = 20 m/s a = 12 m/s2 The...
Read More →Why are mountain roads generally
Question: Why are mountain roads generally made winding upwards rather than going straight up? Solution: The mountain roads are generally made winding upwards rather than going straight up because the force of friction is high which reduces the chances of skidding. When the roads are straight up, the value of frictional force increases which increases the chance of skidding. This is given by: fs = N cos where is the angle of inclination....
Read More →A woman throws an object of mass 500 g
Question: A woman throws an object of mass 500 g with a speed of 25 m/s. (a) what is the impulse imparted to the object? (b) if the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object? Solution: (a) Mass of the object, m = 500 g = 0.5 kg u = 0 v = 25 m/s Impulse is given as: Substituting the values, we get impulse = 12.5 N.s b) m = 0.5 kg u = +25 m/s v = -25/2 m/s ∆p = m(v u) = -18.75 N.s...
Read More →Why does a child feel more pain
Question: Why does a child feel more pain when falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden? Solution: When a child falls on a hard cement floor, the child feels more pain than when the child falls on the soft muddy ground in the garden because the time taken by the child to stop is very small on the cemented ground whereas, on the soft muddy ground, the time taken is more. When the time taken to stop is more, the acceleration is decreased resulti...
Read More →Why are porcelain objects wrapped
Question: Why are porcelain objects wrapped in paper or straw before packing for transportation? Solution: Porcelain objects are wrapped in paper or straw before packing for transportation because they fragile in nature. While transporting these objects change in velocity might bring cracks on them when the acceleration (v u)/t decreases. To avoid the force on these objects, they are wrapped with paper....
Read More →The velocity of a body of mass 2 kg
Question: The velocity of a body of mass 2 kg as a function of t is given by $v(t)=2 t \hat{i}+t^{2} \hat{j}$ Find the momentum and the force acting on it at time $t=2$ sec. Solution: m = 2 kg $\vec{v}(t)=2 t \hat{i}+t^{2} \hat{j}$ $\vec{v}$ at $2 \mathrm{sec}, \vec{v}(2 t)=2(2 t) \hat{i} 2^{2} \hat{j}, v(2)=4 \hat{i}+4 \hat{j}$ Momentum, $\underset{P}{\rightarrow}(2)=m \vec{v}(2)$ $\vec{P}(2)=2[4 \hat{i}+4 \hat{j}], p(2)=8 \hat{i}+8 \hat{j} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$ $\vec{F}=m...
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Question: If $\sin x=\frac{\sqrt{5}}{3}$ and $0x\frac{\pi}{2}$ find the values of sin 2x Solution: Given: $\sin x=\frac{\sqrt{5}}{3}$ To find: $\sin 2 x$ We know that sin2x = 2 sinx cosx (i) Here, we dont have the value of cos x. So, firstly we have to find the value of cosx We know that, $\sin ^{2} x+\cos ^{2} x=1$ Putting the values, we get $\left(\frac{\sqrt{5}}{3}\right)^{2}+\cos ^{2} x=1$ $\Rightarrow \frac{5}{9}+\cos ^{2} x=1$ $\Rightarrow \cos ^{2} x=1-\frac{5}{9}$ $\Rightarrow \cos ^{2} ...
Read More →A person driving a car suddenly applies the brakes
Question: A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing a seat belt, he falls forward and hits his head against the steering wheel. Why? Solution: When a person applies brakes suddenly, he tends to fall forward and hit his head against the steering wheel if he is not wearing the seat belt because his upper body continues to be in a state of motion and in the same direction as that of the car which is due to inertia of motion....
Read More →Prove that
Question: Prove that $\cos 15^{\circ}-\sin 15=\frac{1}{\sqrt{2}}$ Solution: L.H.S $\Rightarrow \cos 15^{\circ}-\sin 15^{\circ}$ $\Rightarrow \cos \left(45^{\circ}-30^{\circ}\right)-\sin \left(45^{\circ}-30^{\circ}\right)$ $\Rightarrow\left(\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}\right)-\left(\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}\right)$ $\Rightarrow\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)-\le...
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