InterviewSolution

Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.	Each of the two guns, mounted on a rotating platform with their lengths parallel to each other, fires 20 bullets per second at a speed of 50 m `s^(-1)`. If the perpendicular distance between the two guns is 1.2 m and the mass of each bullet is 25g, find the couple acting on the platform.
Answer» (i) Force = Impulse `div`Time (ii) Force on the gun = force on the bullets = m `((v-u)/(t))`, where v and u are final and initial velocities of bullets, respectively, and m = mass of each bullet. (iii) Couple = force `xx` perpendicular distance between the forces. (iv) 30 N m

2.	Two wheel and axle systems P and Q are connected as shown in the figure. The radii of wheels and axles of P and Q are 20 cm, 27 cm, 3 cm and 5 cm, respectively. If L = 540 kgwt, find E.
Answer» (i) Mechanical advantage of a wheel and axle (ii) Relate the load of Q to the effort of P. (iii) Mechanical advantage of a wheel and axle is equal to the ratio of the radius of the wheel to that of its axle. (vi) E = 15 kgf

3.	Gears are used in vehicles to transmit motion and power.
Answer» Correct Answer - 1

4.	Find the ratio of efforts required to raise a given load, when the angle of inclination of a given plank is changed from `60^(@)` to `30^(@)`. Also find the percentage change in the mechanical advantage.
Answer» The ratio of efforts `((E_(1))/(E_(2)))` is `(E_(1))/(E_(2))=(mg sin theta_(1))/(mg sin theta_(2)) rArr (E_(1))/(E_(2))=(sin theta_(1))/(sin theta_(2))` Substituting ` sin 60^(@)=(sqrt(3))/(2)` and ` sin 30^(@)=1//2`, we get `(E_(1))/(E_(2))=(sqrt(3))/(2)xx2=sqrt(3)` `:. E_(1):E_(2)= sqrt(3):1` Change in M.A. ` = (1)/(sin 30^(@))-(1)/(sin 60^(@))=2-(2)/(sqrt(3))` `=2(1-(1)/(sqrt(3)))=2((sqrt(3)-1)/(sqrt(3)))` % Change in MA = `("Change in MA")/("initial MA")xx100` `=(2 (sqrt(3)-1))/(sqrt(3)xx(2)/(3))xx100=(sqrt(3)-1)xx100` `= 0.732xx100=73.2%`

5.	The efficiency of a machine is 50% . If 300 J of energy is given to the machine, its output is ________.A. 150 ergB. 350 JC. 250 JD. 150 J
Answer» Correct Answer - D ` eta = ("output")/("input")`=output = ` eta xx ` input `= (50)/(100)xx300J = 150 J`

6.	An effort of 35 N is applied on a machine having mechanical advantage 6. The load that is lifted using the effort is ______ N.A. 210B. 41C. 6D. 29
Answer» Correct Answer - A Load = (effort) `xx` (mechanical advantage) `= (35 N) xx (6) = 210 N`

7.	The pitch of a screw in a screw jack A is half that of another screw jack B. for 10 complete rotations of the lever, which one of the jack will lift a car more higher? Which one need more effort to do an equal work if both the jacks have livers of equal length?
Answer» For one rotation of the lever, the screw moves up by a distance equal to its pitch. `:. ` greater the pitch, higher the car gets lifted. `rArr` Jack B lifts the load higher In thew case of a screw jack, `(W)/(E)=(2 pi L)/(P)` `:. E = (W xx P)/(2 pi L)` As the work done by the two jacks is equal, and the length of their levers is equal, the effort is directly proportional to the pitch, i.e., as the pitch increases, the effort to be applied also increases. Hence, jack B needs more effort.

8.	How can we obtain the resting point-indicating equilibrium, even with unequal masses in the two pans of a physical balance?
Answer» (i) Consider the adjustments made to a physical balance while determining the zero resting point. (ii) By adjusting screws.

9.	The ratio of load to displacement of the rider from its zero mark in a Roman steel yard is 20 gf:1 cm. If the rider is displaced by 20 cm from its zero mark, the load attached to the steel yard is ________.A. 40 gfB. 4 kgfC. 400 gfD. 0.04 kgf
Answer» Correct Answer - C For a load of 20 g , the rider is moved by 1 cm. Thus, for a distance of 20 cm, the load = (20)(20)= 400 gf