# 1) A body of mass m has its position x at a time t expressed by the equation x = 3t^(3/2) + 2t^(-1/2). The instantaneous force on the body is proportional to [i] t^0 [ii] t^(-3/2) [iii] t [iv] t^(3/2 ) 2) A fat hose pipe is held horizontal by a fireman. It delivers water through a nozzle at one litre/sec. If by increasing the pressure, the water is delivered at 2 litre/sec, the fireman now has to: [i] push forward twice as hard [ii] push forward four times as hard [iii] push forward eight times as hard [iv] pull backward four times as hard 3) Is physical independence of force a consquence of first law of motion? If yes, the why is it so? Please reply ASAP. Thank you!

1) A body of mass m has its position *x* at a time *t *expressed by the equation* x = 3t^(3/2) + 2t^(-1/2). *The instantaneous force on the body is proportional to

[i] *t^0* [ii] *t^(-3/2)* [iii]*t *[iv]*t^(3/2*)

2) A fat hose pipe is held horizontal by a fireman. It delivers water through a nozzle at one litre/sec. If by increasing the pressure, the water is delivered at 2 litre/sec, the fireman now has to:

[i] push forward twice as hard

[ii] push forward four times as hard

[iii] push forward eight times as hard

[iv] pull backward four times as hard

3) Is physical independence of force a consquence of first law of motion? If yes, the why is it so?

Please reply ASAP. Thank you!

## 3 Answers

A)In instaneous force Lim (t->0) dP/dt . if u derivate the given equation two times you will get t powers as negative . so after apply limit to the equation force is directly proportional to t^0.

So ans is option A

for instantaneous force

formula is

Lim(t-->0)(m*a)

where a=d^2(x)/dt^2

then if you derivate the given displament equation you wiil get answer negative power of t

and then apply the limit value of t=0

then i.e proportional to t^0

ans : a

Q1. Differentiate twice to get the expression for acceleration as A= 9/4 t^{-1/2} + 3/2 t^{-5/2.} According to the options, none of these is correct.

Q2. Force = Rate of change of momentum = Volume flowing per unit time X density X velocity

and volume flowing per unit time = area of pipe X velocity of water.

so force is propotional to Volume flowing^{2.}

so on doubling the rate of flow, force becomes 4 times.

I can not get your Q3.