Units and Dimensions Multiple choice questions

Multiple choice questions – Units&Dimensions.

(Also Useful for entrance preparation)

1 . 1KWH is unit of
Ans : 1.Time 2. Power 3. Energy 4. Stress
2. Unit of Intensity of magnetic induction field is
Ans : 1.N/Am 2. Tesla   3.Wb/ $m^{2}$     4. All above
3. Which of the following has no units?
Ans : 1. Thermal capacity    2. Magnetic susceptibility 3. Angular acceleration 4. Moment of a magnet
4.Which one of the following units is a fundamental unit?
Ans : 1. watt   2. joule/sec 3. ampere 4. newton
5. $10^5$ Fermi is equal to
Ans : 1. 1 meter      2. 100 micron      3.   1angstrom unit 4. 1 mm
6. kg m/sec is the unit of
Ans : 1. Impulse 2. Angular acceleration 3 . Capacity of condenser   4. Acceleration.
7. candela is the unit of
Ans : 1. Magnetic flux   2. Intensity of electric field    3. Luminous intensity 4. Charge
8. If 10 newton = X dynes, the value of x is
Ans : 1. $10^6$ 2. $10^4$ 3. $10^8$           4. $10^3$
9. 1 KWh is equal to
Ans : 1. 360 J    2. 1800 J   3. $1800times10^5$ J 4. $360times10^5$ J
10. Which of the following is a common unit of a physical quantity in M.K.S & S.I systems.
Ans : 1. ampere 2.kelvin   3. mole 4. joule/sec

11. The fundamental unit which is common in F.P.S and M.K.S systems is
Ans : 1. foot      2. sec 3. kilo gram       4. pound
12. Which of the following is Unit of   length?
Ans 1. Lunar Month      2. Kelvin         3. candela       4. Light year
13. Boltzman’s constant and planck’s constant differ in the dimensions of
Ans : 1. Time and temperature 2. Mass and temperature    3. Length and mass    4. Length and time .
14. Magnetic induction and magnetic flux differ in the dimensions of
Ans : 1 Time   2. Mass 3.Electric current 4.Length
15.Which of the following is a fundamental quantity in M.K.S and C.G.S systems.
Ans: (See other questions)
16 . rad / sec is the unit of
Ans : 1.Angular displacement     2. Angular velocity 3. Angular acceleration   4. Angular momentum .
17 . The ratio of S.I unit of K.E to C.G.S unit of   K.E is
Ans : 1. $10^7$      2. $10^{-7}$      3. $10^{-5}$     4. $10^5$
18. If $u_1$ and $u_2$ are the units of a physical quantity and $n_1$ ‘ $n_1$ are the numerical values, then
Ans : 1. $frac{n_1}{n_2} =frac{u_1}{u_2}$         2. $frac{n_2}{n_1} =frac{u_1}{u_2}$ 3. $frac{n_1}{n_2}^2 =frac{u_1}{u_2}$ 4. None of the above
19. which one of the following is the unit of energy.
Ans : 1. newton   2. N/sec   3. N – sec   4.None of the above
20. Which of the following is not a unit of power .
Ans : 1. Watt     2. joule/hr   3. Nm/sec    4. N/sec
21 . The physical quantity having units of mass is
Ans: 1. Density     2. Momentum       3. Inertia 4. Moment of force
22.Unit used to measure nuclear diameter is
Ans : 1. picometer    2. Fermi 3. Micron     4. millimeter
23. S.I unit of Electric Intensity is
Ans : 1. coulomb / m    2. Henry   3. V / m 4. watt
24. Which is the dimensional formula of the physical quantity whose dimensional S.I unit is Siemen .
Ans : 1. $M^{-1}L^{-2}T^3I^3$    2. $M^{-1}L^{-1}T^3I^3$ 3. $M^1L^{-2}T^3I^3$      4. $M^{-1}L^{-2}T^2I^2$
25. Which is the physical quantity whose dimensional formula is $M^1L^0T^{-2}$ is
Ans : 1. Bulk Modulus 2. Electric conductance    3. Surface tension 4. Frequency
26 . If $M^aL^bT^cI^d$ is the dimensional formula of resistance, then the value of 4a+5b+c-2d = 0
Ans : 1. 15 2. 10            3. 8               4. 12
27. If $M^aL^bT^c$ is the dimensional formula of force, find the value of 2a-b-c
Ans : 1. 8         2. -4     3. 3 4. 6
28. If $M^aL^bT^c$ is the dimensional formula of momentum, and $M^xL^yT^z$ dimensional formula of energy find the value of ax+by-cz .
Ans : 1. -3        2. -1 3. 2       4. 7
29.Which of the following is a derived unit ?
Ans : 1. ampere     2. mole    3. candela    4. newton
30. Which of the following is not a physical quantity ?
Ans : 1. kelvin                2. candela            3. henry      4. all the above
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 amd 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 origional 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 quadrapled, 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.
41)Which of the following is dimensionally correct formula.
a)V=ut + at    b) v+u = at c) V/u =at d) vt = u -a
Soln: The dimensions of L and T are same in all the terms of V+u=at.Hence, according to priciple of homogeneity it a dimensionally correct equation.
42) The dimensional formula of coefficient of kinematic viscosity is
a) $M^0L^2T^{-1}$    b) $M^1L^2T^{-1}$ c) $M^1L^2T^{-3}$ d) $M^0L^3T^{-1}$
43) Which of the following is not the unit of energy?
a) joule   b) Nm   c) W d) $kg m^2sec^{-2}$
44) Dimensional formula of latent heat is
a) $M^1L^1T^{-2}$      b) $1^0L^2T^{-2}$ c) $M^1L^2T^{-1}$       d) $M^0L^2T^{-2}$
45) Pair of quantities having same dimensional formula are
a) velocity , Impulse    b)Force,Weight C)Impulse ,Inertia   d) Angular momtntum, Linear momentum
46) If   R is resistance and L is the inductance,then the dimensions of $frac{R}{L}$ will be same as the dimensions of . . . .
a)time     b) speed    c) frequency d) acceleration
Soln: Dimensional formula of resistance R = $M^1L^2T^{-3}I^{-2}$ and dimensional formula of Inductance is L= $M^1L^2T^{-2}I^{-2}$ .
$frac{R}{L}$ = $frac{M^1L^2T^{-3}I^{-2}} {M^1L^2T^{-2}I^{-2}}$ = $T^{-1}$ which is the dimensional formula of frequency.
47) If the unit of power is 100 erg/minute,the unit of force is 100 dyne and the unit of time is 100 seconds, the unit of length is
a) 5/3 cm b) 2/3 cm c) 1/3 cm d) none
Soln: P = $M^1L^2T^{-3}$ = 100 erg/min =(100/60) erg/sec
F= $M^1L^1T^{-2}$ = 100dyne T= $M^0L^0T^1$ = 100 sec
P= $(M^1L^1T^{-2})(L)(T^{-1})$ = $F(L)(T^{-1})$
L = $PF^{-1}T$ = (100/60)(1/100)(100) =5/3
48 ) If $M^aL^bT^c$ is the dimensional formula of Electric power, find the value of 5a+2b-6c.
a) 25   b) 27 c) 30 d)-9
Soln: Dimensional formula of Electric power = $M^1L^2T^{-3}$ = $M^aL^bT^c$
Comparing the powers of M,L and T we get a=1,b=2 and c=-3 .
Therefore 5a+2b-6c = 5(1)+2(2)-6(-3) =5+4+18 =27.
49) If $M^aL^bT^cI^d$ is the dimensional formula of Magnetic moment, fine the value of a-2b+3c -d
a) 5   b) 7   c)3 d)-5
Soln: Dimensional formula of Magnetic moment is $M^0L^2T^0I^1$ = $M^aL^bT^cI^d$
Comparing the powers of M,L,T and I we get a=0,b=2, c=0 and d=1 ;
Therefore a-2b+3c-d = 0-2(2)+3(0)-1 = -4-1 = -5.
50) If $M^{5a}L^bT^{3c}I^{2d}$ is the dimensional formula of magnetic pole strength, fine the value 5a+b+3c+2d
a) 3    b)2 c) -2   d)0
Dimensional formula of Magnetic pole strength is $M^0L^1T^0I^1$ = $M^{5a}L^bT^{3c}I^{2d}$
Comparing the powers of M,L,T and I we get 5a=0,b=1,3c=0 and 2d=1 ;
Therefore 5a+b+3c+2d = 0+1+0+1 = 2.