Solutions of physics online class test 3 of class 7

 1.What remains constant if a stone falls from a height at a particular place?

Ans. Acceleration due to gravity remains constant at a particular place. It is represented with small 'g'.

g=9.8ms-²

  ≈10ms-²

2.the motion of the wheels of a bicycle is an example of........ type of motion.

Ans. Mixed motion/complex motion.

The motion of The wheels of a bicycles shows two types of motion simultaneously.

Rotatory motion as well as translatory motion.

3.Which one is odd among the following

Ans.A ball thrown by a boy.

Because it shows curvilinear motion.Where as the other threes are examples of rectilinear or linear motion.

4.A man sitting in a moving train is at rest with respect to

Ans.Fellow passengers

5.The the weight of a body is zero 

Ans.At the centre of earth

6. A man completes 4 complete round along the the boundary of a rectangular field 30m long and 20 metre wide in 5minutes & 20 seconds. find his speed?

Ans. Perimeter=2(l+b)

Distance covered in 4 round= 4× 2(l+ b)

  = 4× 2(30m+20m)

= 4×2(50m)

=4×100m

=400m.

Time taken= 5 minutes & 20 seconds.

=5×60s + 20s

= 300s + 20 s

= 320s.

Speed=distance/time.

= 400m/320s

=40m/32s

=5m/4s

=1.25ms-¹.

7.If a body covers 10m, 15m and 20m in three successive seconds. calculate the average speed of the body

Ans.d₁=10m

t₁= 1s

d₂= 15m

t₂=1s

d₃=20m

t₃=1s

Average speed=total distance travelled/total time taken.

=(10m + 15m + 20m ) / (1s+1s+1s)

= 45m / 3s

= 15ms-¹

8.Weight of a box on the the earth surface is 120 Newton .what would be its weight on the Moon surface

Ans. Weight of a body on earth surface is 120 N.

Weight of a body on the moon surface= (1/6)th of weight of box on earth's surface

= 1/6 × 120 N

= 20N

9.A man Travels from Ranchi to Ramgarh with a constant speed of 50 kilometre per hour and returns from Ramgarh to Ranchi with a constant speed of 40 kilometre per hour. find the average speed of the man

Ans. Take the LCM of speed of man.

LCM of 40 & 50 would come as 200.

Take distance as 200km.find t₁& t₂

Speed= distance/time

Speed×time= distance.

Time=distance/speed

t₁= 200km/ 50kmh-¹

   = 4 h

 t₂= 200km/40kmh-¹

    = 5h

Average speed= distance travelled /time taken

Average speed=( 200km + 200km )/ (4h + 5h)

                           = 400km / 9 h

                           = 44.44kmh-¹

                         = 44.44× 1km/h

                        = 44.44 × 5/18 ms-¹

                        =12.3444 ms-¹

10.If a body covers equal distances in an unequal intervals of time the motion is said to be

Ans.non-uniform motion.

11.Which of the following is a periodic motion?

Ans.Pendulum oscillations & Motion of the earth.

So, the correct option is (C). both a and b

12.Weight is a

Ans.Vector quantity.

because weight has both magnitude as well as direction

13.The motion of an object that covers equal distances in equal intervals of time is called 

Ans.Uniform motion

14.Which of the following is not an example of periodic motion

Ans. Mosquito flying in a random direction.