575 - Physics Questions Answers
A body of mass 32kg is suspended by a spring balance from the roof of a vertically operating lift going downward from rest . At that instant the lift covered 20m and 50m , the spring balance showed 30 kg & 36 kg respt. the vel. is increasing at 20m & decreasing at 50m. Plzz sir explain this.
for downward journey
first case
32g-30g = 2g = 32a so a = g/16
second case
32g-36g = -4g = 32a so a = -g/8
so it is acceleration upto 20 m and retardation after that
d benches of a gallery in a cricket stadium r 1m wide & 1m high . a boy strikes d ball at a level of 1m above d ground & hits a sixer. d ball starts at 35m/s at an < of 530 with d horizontal . d benches r perpendicular to plane of motion & d first bench is 110m from d boy. ON WHICH BENCH WILL THE BALL HIT??
ANS 6TH
we know that y = xtanθ - gx2/2u2cos2θ
according to the given conditions
x = 110+n
y = n here n is the number of bench in which ball strikes
θ, u and g are given
A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion of P and R is about
1. 143° 2. 127° 3. 120°
since initially bomb is in rest so initially momentum of the bomb = 0i+0j+0k
after explosion total momentum = 30mi+40mj+mv here v is a vector
now by momentum conservation 0 = 30mi+40mj+mv so v = -30i-40j
so vector along P is i and along R is -3i-4j
so by A.B = |A||B|cosθ
we get i.[-3i-4j] = 1*5*cosθ
so -3/5 = cosθ so θ = 127
If n balls hit elastically and normally on a surface per unit time and all the balls of mass m are moving with same velocity u, then force on surface is
2 mun . How ?
momentum change in elastic collision = mu-(-mu) = 2mu here after collision partical will retrace its direction
for n balls it will be 2mun
so force = rate of change of momentum = 2mun/1 = 2mun
A man of mass 50 kg carries a bag of weight 40 N on his shoulder. The force with which the floor pushes up his feet will be
1. 882 N 2. 530 N 3. 90 N 4. 600 N
reaction by ground is due to both bag and man so 50g+40 = 50*9.8+40 = 490+40 = 530 N
When a 4 kg rifle is fired, the 10 g bullet receives an acceleration of 3 x 10^6 cm/s^2 . The magnitude of the force acting on the rifle (in newton) is
1. 300 2. 200 3. 3000
force on the bullet = [10/1000]*3*104 = 300 N
so that will be same for rifle
A body of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep the body moving with the same velocity is
1. 8 N 2. 0 N 3. 1/2 N
no force is required
Two particles each of mass m are moving in horizontal circle with same angular speed. If both string are of same length then the ratio of tension in string T1 / T2 is
1. 3/2 2. 3 3. 2 4. 1/3
it will be 1:1
A particle is moving on a circular path of radius 1 m with 2 m/s. If speed starts increasing at a rate of 2 m/s² , then acceleration of particle is
centripetal acceleration = v2/r = 2*2/1 = 4
and tangential acceleration = 2 [given]
so acceleration = [42+22]1/2 = √ 20
A particle is revolving in a circular path of radius 2 m with constant angular speed 4 rad/s. The angular acceleration of particle is
1. zero 2. 8π² 3. 16π²
since angular velocity is constant, so zero will be the answer