1. The example of rolling pair is
a) bolt and nut
b) lead screw of a lathe
c) ball and socket joint
d) ball bearing and roller bearing
Explanation: in ball bearing and roller bearing one element rolls over the other element. Hence, they are examples of rolling pair
2. Any point on a link connecting double slider crank chain will trace a
a) straight line
b) circle
c) ellipse
d) parabola
Explanation: One of the inversions of a double slider crank chain is elliptical trammels. So, from the above given options ellipse is best suited.
3.The purpose of a link is to
a) transmit motion
b) guide other links
c) act as a support
d) all of the mentioned
Explanation: all of the mentioned
4. A universal joint is an example of
a) higher pair
b) lower pair
c) rolling pair
d) sliding pair
Explanation: In universal joint, there is surface contact between the two elements. Hence, they form a lower pair.
5. Rectilinear motion of piston is converted into rotary by
a) cross head
b) slider crank
c) connecting rod
d) gudgeon pin
Explanation: In single slider crank chain rotary motion is converted into reciprocating motion
6. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the acceleration.
a) 0.3 m/s2
b) 0.4 m/s2
c) 0.5 m/s2
d) 0.6 m/s2
Explanation: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
202 = 0 + 2a x 500 = 1000a
a = 202/1000 = 0.4 m/s2
7. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the time taken to attain the speed.
a) 50 s
b) 60 s
c) 70 s
d) 80 s
Explanation: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
202 = 0 + 2a x 500 = 1000a
a = 202/1000 = 0.4 m/s2
Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
20 = 0 + 0.4 × t or t = 20/0.4 = 50 s
8. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. If a further acceleration raises the speed to 90 km. p.h. in 10 seconds, find this acceleration and the further distance moved.
a) 0.3 m/s2
b) 0.4 m/s2
c) 0.5 m/s2
d) 0.6 m/s2
Explanation: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
202 = 0 + 2a x 500 = 1000a
a = 202/1000 = 0.4 m/s2
Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
20 = 0 + 0.4 × t or t = 20/0.4 = 50 s
Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : Initial velocity, u = 72 km.p.h. = 20 m/s ;
Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s
Let a = Acceleration of the car.
We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m/s2
9. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. A further acceleration raises the speed to 90 km. p.h. in 10 seconds.The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds. Find the distance travelled during braking.
a) 200 m
b) 300 m
c) 225 m
d) 325 m
Explanation: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
202 = 0 + 2a x 500 = 1000a
a = 202/1000 = 0.4 m/s2
Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
20 = 0 + 0.4 × t or t = 20/0.4 = 50 s
Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : Initial velocity, u = 72 km.p.h. = 20 m/s ;
Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s
Let a = Acceleration of the car.
We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m2
We know that distance moved by the car,
s = ut + 1/2 at2
= 20 x 10 + 1/2 0.5(10)2 = 225 m
10. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?
a) 10.475 rad/s2
b) 11.475 rad/s2
c) 12.475 rad/s2
d) 13.475 rad/s2
Explanation: Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2.
We know that
ω = ω0 + α.t
or 209.5 = 0 + α × 20
α = 209.5 / 20 = 10.475 rad/s2