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Physics >> Mechanics part 1 >> Kinematics Medical Exam

When a force F acts on a particle of mass m, the acceleration of particle becomes a. Now if two forces of magnitude 3F and 4F acts on the particle (both are at an angle of 90° from mass m) simultaneously , then the acceleration of the particle is ..............

 

Asked By: SWATI KAPOOR
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Joshi sir comment

resultant force = 5F

so acceleration = 5a

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A man moves in an open field such that after moving 10m on a straight line, he makes a

sharp turn of 60° to his left. Find the displacement after 8 such turns

Asked By: SWATI KAPOOR
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Joshi sir comment

draw a hexagon

just after 8 such turn displacement = 10√3

Physics >> Mechanics part 1 >> Kinematics Medical Exam

a particle is moving eastwards with a speed of 6m/s. After 6s, the particle is found to be moving with same speed in a direction 60° north of east. The magnitude of average acceleration in this interval of time is ..............

Asked By: SWATI KAPOOR
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Joshi sir comment

initial velocity = 6i

final velocity = 6cos60i+6sin60j

now solve

Physics >> Mechanics part 1 >> Kinematics Medical Exam

a body is dropped from a certain height h (h is very large) and second body is thrown downward with velocity of 5 m/s simultaneously. What will be difference in heights of the two bodies after 3s.

Asked By: SWATI KAPOOR
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Joshi sir comment

g is acting on both bodies so difference = 5*3 = 15 m

Physics >> Mechanics part 1 >> Kinematics Medical Exam

a stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of the tower is .........

Asked By: SWATI KAPOOR
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Joshi sir comment

 (4u)2 = (-u)2+ 2gh

solve

Physics >> Mechanics part 1 >> Kinematics Medical Exam

The velocity of a body depends on time according to the equation v = (t2/10) + 20 . The body is undergoing - 

1. Uniform acceleration

2. Uniform retardation

3. Non-uniform acceleration

Asked By: SWATI KAPOOR
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Joshi sir comment

dv/dt = 2t/10

so a depends on t

option (3)

Physics >> Mechanics part 1 >> Kinematics Medical Exam

When a particle is thrown vertically upwards, its velocity at one third of its maximum height is 10√2 m/s. The maximum height attained by it is......

Asked By: SWATI KAPOOR
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Joshi sir comment

remaining height = 2h/3

so 0= (10√2)- 2g(2h/3)

solve

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A body falling from a vertical height of 10m pierces through a distance of 1m in sand. It faces an average

retardation in sand equal to? 


[Answer is in terms of 'g' i.e acceleration due to gravity ]

Asked By: SWATI KAPOOR
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Joshi sir comment

velocity at the time of impact 

v2 = 02 + 2g10

now in sand this velocity becomes 0 in 1 m

so 02 = 20g - 2a(1) 

so a = 10g

Physics >> Mechanics part 1 >> Kinematics Medical Exam

a body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is...........

(g=10m/s2)

Asked By: SWATI KAPOOR
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Joshi sir comment

from that point to topmost point, time taken = 10/2 = 5 sec

so velocity at that point = 50 m/s

now use

522 = 502- 2gh

solve

A bus starts moving with an acceleration of 2m/s2. A cyclist 96m behind the bus starts simultaneously towards the bus at 20 m/s . After what time cyclist will be able to overtake the bus?

Asked By: NISHANT JAIN
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Joshi sir comment

let after time t

then distance covered by cyclist in this time = 20t

and by bus = 2t2 / 2 = t2

so according to the given condition 

20t = 96+t2

so t = 8,12

after 8 sec cyclist overtakes bus and after 12 sec bus overtakes cyclist

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