A to Z of Physics: Multiple choice questions, Units

31 ) If the unit of length is doubled, unit of time is halved and unit of momentum is quadrupled, the unit of work would change by . . . . . times.

a) 1/8 b) 1/16 c)16 d) 8

soln: Let the original unit of work be W = $M_1L_1^2T_1^{-2}$ ;W= [ $M_1L_1T_1^{-1}$ ][ $L_1$ ][ $T_1^{-1}$ ]

But $p_1$ = $M_1L_1T_1^{-1}$ Therefore W= $p_1L_1T_1^{-1}$ - - - - - - - - - (1)

Let the new units of work, momentum, length and time be W’,p’,L’ and T’ respectively.

Given that p’=4 $p_1$ ; L’ = 2 $L_1$ and T’= (1/2) $T_1$

The new unit of work W’ = $p'L'T'^{-1}$ ;

Substituting the values of p’,L’ and T’ in above equation we get W’ = $(4p_1)(2L_1)(\frac{T_1}{2})^{-1}$

W’ = 16 $p_1L_1T_1^{-1}$ = 16W .

32 ) If the unit of force were 20N,that of power were 1MW and that of time were 1 millisecond then the unit of length would be

a) 20m b)50m c) 100m d) 1000m

Soln: Given that unit of power P=[ $ML^2T^{-3}$ ] = $10^6$ W; F= [ $MLT^{-2}$ ]=20N;T= $10{-3}$ sec

P= $MLT^{-2}LT^{-1}$ = $FLT^{-1}$

Substitute the values of P,F and T in above equation $10^6$ = 20(L) $(10^{-1})^{-3}$

L= $\frac{10^6}{{20}\times{10^3}}$ = 50m

33 ) If the unit of force is 10N,that of length is 2m and that of velocity is 100m/sec, then the unit of mass is

a)0.002kg b) 2kg c ) 20kg d) 0.2kg

Soln : Given that unit of F=10N = [ $MLT^{-2}$ ]; unit of length L= 2m ; unit of velocity V=100m/sec=[ $LT^{-1}$

The unit of F=[ $MLT^{-2}$ ] = [ $M$ ][ $(LT^{-1})^2$ ] $\div$ [L]

F= $\frac{MV^2}{L}$ ; M = $\frac{FL}{V^2}$

M = $\frac{(10)(2)}{100^2}$ = $\frac{20}{10000}$ = 0.002 kg

34) A force 100N acts on a body.If the units of mass and length are doubled and unit of time is halved,then the force in the new system changes to

a)160N b) 1.6 N c) 16N d) 1600N

Soln: Let the original unit of force F=100N=[ $MLT^{-2}$ ]

Let the new unit of force,length,mass and time be F’, L’, M’ and T’ respectively.

Given that units of mass and length are doubled i.e L’=2L and M’=2M and unit of time is halved T’=T/2

The new unit of force F’=[ $M'L'T'^{-2}$ ] = [ $(2M)(2L)(\frac{T}{2})^{-2}$ = 16 [ $MLT^{-2}$ ]=16F =1600N

35) The unit of energy is 10J,if the unit of mass is tripled,unit of acceleration is doubled and the unit of length is halved. What will be the new unit of energy.

a)15J b) 30J c) 300J d) 3J

Soln: Original unit of power is E= $ML^2T^{-2}$ =10J

E= $ML^2T^{-2}$ = $(M)(LT^{-2})(L)$ = $(M)(a)(L)$

Let the new units of mass, acceleration and length be M’,a’ and L’ respectively

Given that L’= L/2 , a’=2a and M’=3M

The new unit of energy E = $M'L'^2T'^{-2}$ = $(M')(L'T'^{-2})(L)$ = $(M')(a')(L')$

E = $(3M)(2a)(L/2)$ = 3 $(M)(a)(L)$ = 3E = 30J

36) The power of a moter is 150W.If the unit of force is doubled,unit of velocity is tripled what will be the new unit of power.

a) 600W b)750W c) 900W d) 300w

Soln: The original unit power of is P=150W=[ $ML^2T^{-3}$ ]

The Unit of force F=[ $MLT^{-2}$ ] and unit of velocity V=[ $LT^{-1}$ ]

P=[ $(MLT^{-2}$ ][ $LT^{-1}$ ] = F V

Let the new unit of power,force and velocity be P’,F’ and V’ respectively.

Given that F’ = 2F and V’ = 3V

P’ = [ $M'L'^2T'^{-3}$ ]= [ $(M'L'T'^{-2})(L'T'^{-1})$ ]=F’V’

P’ = (2F)(3V) = 6FV = 6P =6(150) = 900W

37) The electric resistance of a conductor is 54 ohm.If the unit of mass and length are tripled, units of time and electric current are doubled.Then the value of new electric resistance.

a)540 ohm b) 1080 ohm c) 1620 ohm d)1944 ohm

Soln: The original unit of electric resistance R =[[ $ML^2T^{-3}I^{-2}$ ]=54 ohm

Let the changed units of resistance,length,mass,time and current are R’,L’,M’,T’ and I’ respectively.

Given that L’=3L , M’ =3M , T’ = 2T and I’=2I

R’ = [[ $M'L'^2T'^{-3}I'^{-2}$ ] ; R’ =[[ $(3M)(3L)^2(2T)^{-3}(2I)^{-2}$ ]

R’ = 36[[ $ML^2T^{-3}I^{-2}$ ] = 36R =36(54) = 1944 ohm.

38)The unit of angular momentum is 25 $kg.m^2sec^{-1}$ . If the momentum is doubled and length is quadrupled, what will be the new unit of angular momentum.

a)200units b) 150 units c) 400 units d)600units

Soln: let the angular momentum X = [ $ML^2T^{-1}$ ] = 25

momentum p =[ $MLT^{-1}$ ]

X =[[ $(MLT^{-1})(L)$ ] ; X = p (L)

let the new Angular momentum, momentum and length be X’,p’ and L’ respectively.

Given that p’ = 2p and L’ = 4L

X’ = [ $M'L'^2T'^{-1}$ ] = [ $(M'L'T'^{-1})(L')$ ]

X’ = p’ (L’) = (2p)(4L) = 8pL =8 X = 8(25) = 200 units

39) The ratio of C.G.S unit of gravitational constant to S.I unit is

a) $10^2$ b) $10^3$ c) $10^{-2}$ d) $10^{-3}$

Soln: Let the S.I unit of gravitational constant G =[ $M^{-1}L^3T^{-2}$ ] and C.G.S unit be G’ =[ $M'^{-1}L'^3T'^{-2}$ ] ;

$\frac{G'}{G}$ = $\frac{M'^{-1}L'^3T'^{-2}}{M^{-1}L^3T^{-2}}$

M = 1000 M’ ; L= 100L’ and T = T’

Substituti0n these values in above equation we get $\frac{G'}{G}$ = $\frac{M'^{-1}L'^3T'^{-2}}{(1000M')^{-1}(100L')^3T^{-2}}$

$\frac{G'}{G}$ = $\frac{1000}{10^6}$ = $10^{-3}$

40 ) If the units of mass,length and time are doubled how does the unit of coefficient of linear expansion will changes

a) becomes 8times b) becomes 16 times c) becomes 1/2 d) remains same

Soln: Coefficient of linear expansion $\alpha$ = [ $M^0L^0T^0K^{-1}$ ] .From the dimensional formula it is clear that it doen not depend on mass,length and time.even if the these physical quantities changes the Coefficient of linear expansion will remain same.