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.

A student measures the time period of `100` ocillations of a simple pendulum four times. The data set is `90 s`, 91 s, 95 s, and 92 s`. If the minimum division in the measuring clock is `1 s`, then the reported men time should be:A. (a) `92 +- 1.8 s`B. (b) `92 +- 3 s`C. ( c ) `92 +- 2 s`D. ( d ) `92 +- 5.0 s`

Answer» Correct Answer - A
(a) `DeltaT = (| DeltaT_(1)| + |DeltaT_(2)| + DeltaT_(3)| + |DltaT_(4))|/(4)`
`= (2+1+3+0)/(4) = 1.5`
As the resolution of measuring clock is `1.5` therefore the mean time should be `92+-1.5`
Discussion
2.

in an experiment the angles are required to be using an instrument, `29` divisions of the main scale exactly coincide with the `30` divisions of the vernier scale. If the sallest division of the main scale is half- a degree `(= 0.5^(@)`, then the least count of the instrument is :A. (a) half minuteB. (b) one degreeC. ( c ) half degreeD. (d) one minute

Answer» Correct Answer - D
(d) `30 Division sof vernier scale coincide with `29` divisions of main scales
Therefore `1 V.S.D = (29)/(30)MSD`
Least count = 1MSD - 1VSD = 1MSD - (29)/(30)MSD`
`=(1)/(30)/(MSD) = (1)/(30) xx [email protected] = 1minute`.
Discussion
3.

In a screw gauge, the zero of mainscale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are `50`. It moves `0.5 mm` on main scale In one rotation. The diameter of the ball in figure (ii) is A. (a) `2.25mm`B. (b) `2.20mm`C. ( c ) `1.20mm`D. (d) `1.25mm`

Answer» Correct Answer - C
( c ) `Least count = (0.5)/(50) = 0.01mm`
Zero error `= 5 xx L.C = 5 xx 0.01mm = 0.05mm`
`Diameter of ball = [Reading on main scale] + [Reading on circular scale xx L.C] - Zero error`
`= 0.5 xx 2 + 25 xx 0.01 - 0.05 = 1.20mm`
Discussion
4.

A student performs an experiment an for determination of `g(=(4pi^(2)l)/(T^(2)))`. The error in length `l` is `Deltal` and in time `T` is `DeltaT and n` is number of times the reading is taken. The measurment of `g` is most accurate forA. (a) `Deltal = 5mm, Delta =0.2sec, n = 10`B. (b) `Deltal = 5mm, Delta =0.2sec, n = 20`C. ( c ) `Deltal = 5mm, Delta =0.1sec, n = 10`D. ( d ) `Deltal = 1mm, Delta =0.1sec, n = 50`

Answer» Correct Answer - D
(d) `(Deltag)/(g) = (Deltal)/(l)+2(DeltaT)/(T)`
`Deltal and DeltaT` least and number of reading are maximum in option (d), therefore the measurement of `g` is most accurate with data used in this option.
Discussion
5.

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : `58.5 degree` Vernier scale reading : `09` divisions Given that `1` division on main scale correspods to `0.5` degree. Total divisions on the vernier scale is `30` and match with `29` divisions of the main scale. the angle of the prism from the above data:A. (a) `58.59 degree`B. (b) `58.77 degree`C. ( c ) `58.65 degree`D. (d) `59 degree`

Answer» Correct Answer - C
( c ) `because` Reading of vernier = Main scale reading + Vernier scale reading xx least count`.
Main scale reading = 58.5`
`Vernier scale reading = 09 division`
`least count of Vernier = [email protected]//30`
Thus `R = [email protected] + 9 xx ([email protected])/(30)`
`R = 58.65`
Discussion
6.

There are two Vernier calipers both of which have `1cm` divided into `10` equal divisions on the main scale. The vernier scale of the calipers `( c_(1))` has `10` equal divisions that correspond to `9` main scale divisions. The Vernier scale of the other calipers `( C_(2))` has `10` equal divisions that correspond to `11` main scale divisions. the reading of the two calipers are shown in the figure. the measured values (in cm) by calipers `C_(1) and C_(2)` respectively, are A. (a) `2.85 and 2.82`B. (b) `2.87 and 2.83`C. ( c ) `2.87 and 2.86`D. (d) `2.87 and 2.87`

Answer» Correct Answer - B

For`C_(2)
`LgtC =1mm-1.1mm` [10VSD = 11mm]`
`L.C =-0.1mm = 0.01cm`
Reading = 2.8 + (10-7)xx0.01 = 2.83 cm`
Discussion
7.

A vernier calipers has `1mm `marks on the main scale. It has `20` equal divisions on the Verier scale which match with `16` main scale divisions. For this Vernier calipers, the least count isA. (a) `0.02mm`B. (b) `0.05mm`C. ( c ) `0.1mm`D. (d) `0.2mm`

Answer» Correct Answer - D
(d) `20` division on the vernier scale
`=16` divisions of main scale
:. `1` division on the vernier scale
`=(16)/(20) divisions of main sacle = (16)/(20 xx 1mm = 0.8mm`
We knoe that least count = 1MSD - 1VSD`
`=1mm - 0.8mm = 0.2 mm`
Discussion
8.

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between `5.10cm and 5.15cm` of the main scale. The Vernier scale has `50` divisions equivalent to `2.45 cm`. The `24^(th)` division of the Vernier scale exactly coincides with one of the main scale divisions. the diameter of the cylinder isA. (a) `5.112cm`B. (b) `5.124cm`C. ( c ) `5.136cm`D. (d ) `5.1148cm`

Answer» Correct Answer - B
(b) Reading = M.S.R + No of division of `V.S matching the main scale division `(1MSD -1VSD)`
`= 5.10 + 24((0.05 -(2.45)/(50))`
`=5.124cm` option (b) is correct.
Discussion
9.

A wire of length `l = +- 0.06cm` and radius `r =0.5+-0.005 cm` and mass `m = +-0.003gm`. Maximum percentage error in density isA. (a) `4`B. (b) `2`C. ( c ) `1`D. (d) `6.8`

Answer» Correct Answer - A
(a) `rho = (m)/(l pir^(2))`
`(Deltarho)/(rho) = (Deltam)/(m) + (2pir)/( r ) + (Deltal )/(l )`
Putting the values
`Deltal = 0.06 cm, l = 6cm`, `Deltar = 0.005 cm`, `r = 0.5cm`,
`m= 0.3gm`, `Deltam = 0.003gm`
:. `(Deltarho)/(rho) = (4)/(100)`
:. `(Deltarho)/(rho) xx 100 = 4%`
Discussion
10.

Write the dimensions of the following in terms of mass, time, length and charge (i) magnetic flux (ii) rigidity modulus

Answer» Correct Answer - A::B
`Meganetic Flux = [M^(1)L^(2)T^(-1)Q^(-1)]`
Modulus of Rigidity = [ML^(1)T^(-2)]`
Discussion
11.

Identify the pair whose dimensions are equalA. (a) torque and WorkB. (b) stress and energyC. ( c ) force and stressD. (d) force and work

Answer» Correct Answer - A
(a) `W = vec(F).vec(s) = Fs cos theta`
`=[MLT^(-2)][L] = [ML^(2)T^(-2)]`,
`vec(tau) = vec( r ) xx vec(F)` rArr `tau = rF sin theta`
`= [L][MLT^(-2)] = [ML^(2)T^(-2)]`
Discussion
12.

In terms of potential difference `C`, electric current`I` , permittivity `epsilon_(0)`, permeability `mu_(0)` and speed of light `c`, the dimensionally correct equation `(s)` is `(are)`A. (a) `mu_(0)I^(2) = epsilon_(0)V^(2)`B. (b) `mu_(0)I = mu_(0)V`C. ( c ) `I = epsilon_(0)cV`D. (d) `mu_(0)cI = epsilon_(0)V`

Answer» Correct Answer - A::C
(a, c) We know that
`C = (1)/(sqrtmu_(0)epsilon_(0)) and = R sqrt((mu_(0))/(epsilon_(0))`
Now, `mu_(0)I^(2) = epsilon_(0)V^(2)`
:. `(mu_(0))/(epsilon_(0)) = (V^(2))/(I^(2)) = R^(2)` rArr Option `A` is correct
Now,`epsilon)(0)I = mu_(0)V`
:. `(mu_(0))/(epsilon_(0)) = (I)/(V) = (I)/(R)` rArr option `B` is incorrect
Now, `I = epsilin_(0)CV`
:. `(1)/(epsilon_(0)C) = (V)/(I) = R`
:. `(1)/(epsilon_(0)(1)/(sqrtmu_(0)epsilon_(0)) = R`
:. `sqrt((mu_(0))/(epsilon_(0))) = R` rArr Option `C` is correct
Now, `mu_(0)cI = epsilon_(0)V`
:. `(mu_(0))/(epsilon_(0)) = (V)/(I_(c)) = (R)/(C) = sqrt((mu_(0))/(epsilon_(0))) xx ((1)/(1))/sqrt(mu_(0)epsilon_(0)) = mu_(0)`
rArr Option `(d)` is incorrect
Discussion
13.

Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of the vacuum, and `[mu_(0)]` that of the permeability of the vacuum. If `M = mass ,L = length, T = time and I = electric current`,A. (a) `[epsilon_(0)] = M^(-1)l^(-3)T^(2)I`B. (b) `[epsilon_(0)] = M^(-1)l^(-3)T^(4)I^(2)`C. ( c ) `[mu_(0)] = MLT^(-2)I^(-2)`D. (d) `[mu_(0)] = ML^(2)T^(-1)I`

Answer» Correct Answer - B::C
(b,c) By defination `F = (Q_(1)Q_(2))/((4piepsilon_(0))r^(2)) and (F)/(l) = (mu_(0)I_(1)I_(2))/(2piL)`
Hence,`[epsilon_(0)] = [Q^(2)]/([F][r^(2)]) = (I^(2)T^(2))/(MLT^(-2)L^(2)) = M^(-1)L^(-3)T^(4)I^(2)`
`[mu_(0)] = ([F])/([I]^(2)) = (MLT^(-2))/(I^(2)) = ML^(-2)T^(-2)`
Discussion
14.

The dimension of `((1)/(2))epsilon_(0)E^(2)` (`epsilon_(0)` : permittivity of free space, E electric fieldA. (a) `MLT^(-1)`B. (b) `ML^(2)T^(2)`C. ( c ) `ML^(-1)T^(-2)`D. ( d ) `ML^(2)T^(-1)`

Answer» Correct Answer - C
( c ) Note : here `((1)/(2))epsilon_(0)E^(2)` represents energy per unit volume.
`[epsilon_(0)][E^(2)] = ([En ergy])/([Volume]) = (ML^(2)T^(-2))/(L^(3)) = ML^(-1)T^(-2)`
Discussion
15.

The dimension of magnetic field in `M,L,T and C` (coulomb) is given asA. (a) `MLT^(-1)C^(-1)`B. (b) `MT^(2)C^(-2)`C. ( c ) `MT^(-1)C^(-1)`D. (d) `MT^(-2)C^(-1)`

Answer» Correct Answer - C
( c ) we know that `F = q v B`
:. `B = (F)/(qv) = (MLT^(-2)/(C xx LT^(-1)) = MT^(-1)C^(-1)`
Discussion
16.

Which one of the following represents the correct dimensions of the coefficient of viscosity?A. (a) `ML^(-1)T^(-1)`B. (b) `MLT^(-1)`C. ( c ) `ML^(-1)T^(-2)`D. (d) `ML^(-2)T^(-2)`

Answer» Correct Answer - A
(a) From stokes law `F = 6pi etarv rArr eta = (F)/(6pirv)`
:. `eta = (MLT^(-2))/([L][LT^(-1)])` rArr eta = [ML^(-1)T^(-1)]`
Discussion