A person of mass 50 kg stands on a weighing
Question: A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m/s2 what would be the reading of the weighing scale? Solution: The apparent weight on the weighing scale decreases when the lift descends with an acceleration a. Let W be the apparent weight Therefore, W = R = (mg ma) = m(g a) Therefore, apparent weight, W = 50(10-9) = 50 N Reading of the weighing scale = R/g = 50/10 = 5 kg....
Read More →A girl riding a bicycle along a straight road
Question: A girl riding a bicycle along a straight road with a speed of 5 m/s throws a stone of mass 0.5 kg which has a speed of 15 m/s with respect to the ground along her direction of motion. The mass of the girl and the bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so? Solution: Given, m1= 50 kg m2= 0.5 kg u1= 5 m/s u2= 5 m/s v1= ? v2= 15 m/s From the law of conservation of momentum, Initial momentum = final momentum (m1+ m2)...
Read More →A body of mass 10 kg is acted upon
Question: A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is (a) 1 m/s2at an angle of tan-1(4/3) w.r.t 6N force (b) 0.2 m/s2at an angle of tan-1(4/3) w.r.t 6N force (c) 1 m/s2at an angle of tan-1(3/4) w.r.t 8N force (d) 0.2 m/s2at an angle of tan-1(3/4) w.r.t 8N force Solution: The correct answers are (a) 1 m/s2at an angle of tan-1(4/3) w.r.t 6N force (c) 1 m/s2at an angle of tan-1(3/4) w.r.t 8N force...
Read More →Two billiard balls A and B, each of mass 50 g
Question: Two billiard balls A and B, each of mass 50 g and moving in opposite directions with speed of 5 m/s each, collide and rebound with the same speed. If the collision lasts for 10-3seconds, which of the following statements are true? (a) the impulse imparted to each ball is 0.25 kg.m/s and the force on each ball is 250N (b) the impulse imparted to each ball is 0.25 kg.m/s and the force exerted on each ball is 25 10-5N (c) the impulse imparted to each ball is 0.5 Ns (d) the impulse and the...
Read More →The motion of a particle of mass m is given
Question: The motion of a particle of mass m is given by x = 0 for t 0 sec, x(t) = A sin 4p t for 0 t (1/4) sec, and x = 0 for t (1/4) sec. Which of the following statements is true? (a) the force at t = (1/8) sec on the particle is -162Am (b) the particle is acted upon by on impulse of magnitude 42Am at t = 0 sec and t = (1/4) sec (c) the particle is not acted upon by any force (d) the particle is not acted upon by a constant force (e) there is no impulse acting on the particle Solution: The co...
Read More →Prove that
Question: Prove that $2 \sin 75^{\circ} \sin 15^{0}=\frac{1}{2}$ Solution: L.H.S $=2 \sin 75^{\circ} \sin 15^{\circ}$ $=2 \sin \left(45^{\circ}+30^{\circ}\right) \sin \left(45^{\circ}-30^{\circ}\right)$ $=\cos \left(45^{\circ}-30^{\circ}-45^{\circ}-30^{\circ}\right)-\cos \left(45^{\circ}+30^{\circ}+45^{\circ}-30^{\circ}\right)$ $=\cos \left(-60^{\circ}\right)-\cos 90^{\circ}$ $=\cos 60^{\circ}-0$ $=\frac{1}{2}$...
Read More →A car of mass m starts from
Question: A car of mass m starts from rest and acquires a velocity along the east $v=v \hat{i}$ in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the caris (a) mv/2 eastward and is exerted by the car engine (b) mv/2 eastward and is due to the friction on the tyres exerted by the road (c) more than mv/2 eastward exerted due to the engine and overcomes the friction of the road (d) mv/2 exerted by the engine Solution: The correct answer is (b) mv/2 eastward and ...
Read More →A body with mass 5 kg
Question: A body with mass 5 kg is acted upon by a force $F=(-3 \hat{i}+4 \hat{j})$ N. If its initial velocity at $\mathrm{t}=0$ is $v=(6 \hat{i}-12 \hat{j}) \mathrm{m} / \mathrm{s}$, the time at which it will just have a velocity along the $\mathrm{y}$-axis is (a) never (b) 10 s (c) 2 s (d) 15 s Solution: The correct answer is (b) 10 s...
Read More →A body of mass 2 kg travels according to the law
Question: A body of mass 2 kg travels according to the law x(t) = p(t) + qt2 + rt3 where p = 3 m/s, q = 4 m/s2, and r = 5 m/s3. The force acting on the body at t=2 seconds is (a) 136 N (b) 134 N (c) 158 N (d) 68 N Solution: The correct answer is a) 136 N...
Read More →A hockey player is moving northward
Question: A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is (a) frictional force along westward (b) muscle force along southward (c) frictional force along south-west (d) muscle force along south-west Solution: The correct answer is (c) frictional force along the south-west...
Read More →Verify Rolle's theorem for each of the following functions on the indicated intervals
Question: Verify Rolle's theorem for each of the following functions on the indicated intervals (i) $f(x)=x^{2}-8 x+12$ on $[2,6]$ (ii) $f(x)=x^{2}-4 x+3$ on $[1,3]$ (iii) $f(x)=(x-1)(x-2)^{2}$ on $[1,2]$ (iv) $f(x)=x(x-1)^{2}$ on $[0,1]$ (v) $f(x)=\left(x^{2}-1\right)(x-2)$ on $[-1,2]$ (vi) $f(x)=x(x-4)^{2}$ on the interval $[0,4]$ (vii) $f(x)=x(x-2)^{2}$ on the interval $[0,2]$ (viii) $f(x)=x^{2}+5 x+6$ on the interval $[-3,-2]$ Solution: (i) Given: $f(x)=x^{2}-8 x+12$ We know that a polynomia...
Read More →Conservation of momentum in a collision
Question: Conservation of momentum in a collision between particles can be understood from (a) conservation of energy (b) Newtons first law only (c) Newtons second law only (d) both Newtons second and third law Solution: The correct answer is (d) both Newtons second and third law...
Read More →Prove that
Question: Prove that $2 \cos 45^{\circ} \cos 15^{\circ}=\frac{\sqrt{3}+1}{2}$ Solution: L.H.S $=2 \cos 45^{\circ} \cos 15^{\circ}$ $=2 \cos 45^{\circ} \cos \left(45^{\circ}-30^{\circ}\right)$ $=2 \frac{1}{\sqrt{2}}\left(\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}\right)$ $=\sqrt{2}\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)$ $=\sqrt{2}\left(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)$ $=\sqrt{2}\left(\frac{\sqrt{...
Read More →Verify Rolle's theorem for each of the following functions on the indicated intervals
Question: Verify Rolle's theorem for each of the following functions on the indicated intervals (i) $f(x)=x^{2}-8 x+12$ on $[2,6]$ (ii) $f(x)=x^{2}-4 x+3$ on $[1,3]$ (iii) $f(x)=(x-1)(x-2)^{2}$ on $[1,2]$ (iv) $f(x)=x(x-1)^{2}$ on $[0,1]$ (v) $f(x)=\left(x^{2}-1\right)(x-2)$ on $[-1,2]$ (vi) $f(x)=x(x-4)^{2}$ on the interval $[0,4]$ (vii) $f(x)=x(x-2)^{2}$ on the interval $[0,2]$ (viii) $f(x)=x^{2}+5 x+6$ on the interval $[-3,-2]$ Solution: (i) Given: $f(x)=x^{2}-8 x+12$ We know that a polynomia...
Read More →In the previous problem,
Question: In the previous problem, the magnitude of the momentum transferred during the hit is (a) zero (b) 0.75 kg.m/s (c) 1.5 kg.m/s (d) 14 kg.m/s Solution: The correct option is (c) 1.5 kg.m/s...
Read More →A cricket ball of mass
Question: A cricket ball of mass $150 \mathrm{~g}$ has an initial velocity $u=(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}$ and a final velocity $v=-(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}$ after being hit. The change in momentum (a) zero (b) $-(0.45 \hat{i}+0.6 \hat{j})$ (c) $-(0.9 \hat{i}+1.2 \hat{j})$ (d) $-5(\hat{i}+\hat{j})$ Solution: The correct answer is $\mathrm({c})-(0.9 \hat{i}+1.2 \hat{j})$...
Read More →Solve this
Question: If $\cos x+\cos y=\frac{1}{3}$ and $\sin x+\sin y=\frac{1}{4}$, prove that $\tan \left(\frac{x+y}{2}\right)=\frac{3}{4}$ Solution: $\cos x+\cos y=\frac{1}{3}$ .............$-i$ $\sin x+\sin y=\frac{1}{4}$ ................-ii dividing ii by I we get, $\Rightarrow \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{\frac{1}{4}}{\frac{1}{3}}$ $\Rightarrow \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{3}{4}$ $\Rightarrow \frac{2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \cos \frac{x-y}...
Read More →A metre scale is moving with uniform velocity.
Question: A metre scale is moving with uniform velocity. This implied (a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale (b) the force acting on the scale is zero and the torque acting about the centre of mass of the scale is also zero (c) the total force acting on it need not be zero but the torque on it is zero (d) neither the force nor the torque needs to be zero Solution: The correct answer is (b) the force acting on the scale is zero and th...
Read More →A ball is travelling with uniform translator motion.
Question: A ball is travelling with uniform translator motion. This means that (a) it is at rest (b) the path can be a straight line or circular and the ball travels with uniform speed (c) all parts of the ball have the same velocity and the velocity is constant (d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly Solution: The correct option is (c) all parts of the ball have the same velocity and the velocity is constant...
Read More →Prove that
Question: Prove that $\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}=\frac{3}{16}$ Solution: L.H.S $=\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}$ $=\frac{1}{2}\left(2 \cos 70^{\circ} \cos 10^{\circ}\right) \cos 50^{\circ} \frac{\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{4}\left\{\cos \left(70^{\circ}+10^{\circ}\right)+\cos \left(70^{\circ}-10^{\circ}\right)\right\} \cos 50^{\circ}$ $=\frac{\sqrt{3}}{4}\left\{\cos 80^{\circ} \cos 50^{\circ}+\cos 60^{\circ} \cos 50^{\circ...
Read More →Prove that
Question: Prove that $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{3}{16}$ Solution: L.H.S $=\frac{1}{2}\left(2 \sin 80^{\circ} \sin 20^{\circ}\right) \sin 40^{\circ} \frac{\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{4}\left\{\cos \left(80^{\circ}-20^{\circ}\right)-\cos \left(80^{\circ}+20^{\circ}\right)\right\} \sin 40^{\circ}$ $=\frac{\sqrt{3}}{4}\left\{\cos 60^{\circ} \sin 40^{\circ}-\cos 100^{\circ} \sin 40^{\circ}\right\}$ $\left.=\frac{\sqrt{3}}{4}\left\{\frac{1}{2} \sin 40^{\...
Read More →Prove that
Question: Prove that $\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}=\frac{1}{16}$ Solution: L.H.S $=\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}$ $=\frac{1}{2}\left(2 \sin 70^{\circ} \sin 10^{\circ}\right) \sin 50^{\circ} \frac{1}{2}$ $=\frac{1}{4}\left\{\cos \left(70^{\circ}-10^{\circ}\right)-\cos \left(70^{\circ}+10^{\circ}\right)\right\} \sin 50^{\circ}$ $=\frac{1}{4}\left\{\cos 60^{\circ} \sin 50^{\circ}-\cos 80^{\circ} \sin 50^{\circ}\right\}$ $\left.=\f...
Read More →Prove that
Question: Prove that $\frac{\cos 2 x \sin x+\cos 6 x \sin 3 x}{\sin 2 x \sin x+\sin 6 x \sin 3 x}=\cot 5 x$ Solution: $=\frac{\cos 2 x \sin x+\cos 6 x \sin 3 x}{\sin 2 x \sin x+\sin 6 x \sin 3 x}$ $=\frac{2 \cos 2 x \sin x+2 \cos 6 x \sin 3 x}{2 \sin 2 x \sin x+2 \sin 6 x \sin 3 x}$ $=\frac{\sin (2 x+x)-\sin (2 x-x)+\{\sin (6 x+3 x)-\sin (6 x-3 x)\}}{\cos (2 x-x)-\cos (2 x+x)+\cos (6 x-3 x)-\cos (6 x+3 x)}$ $=\frac{\sin 3 x-\sin x+\sin 9 x-\sin 3 x}{\cos x-\cos 3 x+\cos 3 x-\cos 9 x}$ $=\frac{\s...
Read More →A hill is 500 m high. Supplies are to be sent across
Question: A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of hill and can be moved on the ground at a speed of 2 m/s so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g = 10 m/s2. Solution: Given, Speed of packets = 125 m/s Height of the hill = 500 m Distance bet...
Read More →|A|=2 and |B|=4, then match the relations in
Question: |A|=2 and |B|=4, then match the relations in column I with the angle $\theta$ between $\mathrm{A}$ and $\mathrm{B}$ in column II. Solution: (a) matches with (iv) (b) matches with (iii) (c) matches with (i) (d) matches with (ii)...
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