Science > Physics > Units and Measurement > JEE Main/AIEEE Past Questions: Units and Measurements
MCQs With One Correct Answer:
Choose the Correct Alternative
Q1. Identify the pair whose dimensions are equal (2002)
(a) torque and work
(b) stress and energy
(c) force and stress
(d) force and work
Q2. Dimensions of 1/μoεo, where symbols have their usual meanings, are (2003)
(a) [L-1 T]
(b) [L-2T2]
(c) [L2T-2]
(d) [LT-1]
3. The physical quantities not having same dimensions are (2003)
(a) torque and work
(b) momentum and Planck’s constant
(c) stress and Young’s modulus
(d) speed and (μoεo)-1/2
Q4. Which of the following represents the correct dimensions of the coefficient of viscosity? (2004)
(a) [ML-1T-1]
(b) [MLT-1]
(c) [ML-1T-2]
(d) [ML-2T-2]
Q5. Out of the following pair, which one does not have identical dimensions is (2005)
(a) Impulse and momentum
(b) angular momentum and Planck’s constant
(c) work and torque
(d) moment of inertia and moment of force (towards north-west)
Q6. The dimensions of magnetic field in M, L, T and C (coulomb) is given as (2008)
(a) [MLT-1C-1]
(b) [MT2C-2]
(c) [MT-1C-1]
(d) [MT-2C-1]
Q7. A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms-1. The magnitude of ots momentum is recorded as (2008)
(a) 17.6 kg ms-1
(b) 17.565 kg ms-1
(c) 17.56 kg ms-1
(d) 17.57 kg ms-1
Q8. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scal as 35. The diameter of wire is (2008)
(a) 3.32 mm
(b) 3.73 mm
(c) 3.67 mm
(d) 3.38 mm
Q9. In an experiment the angles are required to be measured using an instrument, 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half a degree (0.5o) then the least count of the instrument is (2009)
(a) half minute
(b) one degree
(c) half degree
(d) one minute
Q10. The respective numbers of significant figures for the numbers 23.023, 0.0003, and 2.1 x 10-3 are (2010)
(a) 5, 1, 2
(b) 5, 1, 5
(c) 5, 5, 2
(d) 4, 4, 2
Q11. A screw gague gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm and Circular scale reading : 52 divisions. Given that 1 mmm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is (2011)
(a) 0.052 cm
(b) 0,026 cm
(c) 0.005 cm
(d) 0.52 cm
Q12. Resistance of a given wire is obtained by measuring the current flowing in it and voltage difference aplied across it. If the percentage error in mthe measurement of the current and voltage difference are 3% each, then the error in value of resistance of the wire is (2012)
(a) 6%
(b) zero
(c) 1%
(d) 3%
Q13. A spectrometer gives following reading when used to measure the angle of prism. Main scale reading: 58.5 degree, Vernier scale reading: 09 divisions. Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the Vernier scale is 30 and match with 30 divisions of the main scale. The angle of prism from the above data is (2012)
(a) 58.59 degree
(b) 58.77 degree
(c) 58.65 degree
(d) 59 degree
Q14. Let [e0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = Electric current, then (JEE Main 2013)
(a) [M-1L-3T2A]
(b) [M-1L-3T4A2]
(c) [M1L2T1A2]
(d) [M1L2T1A]
Q15. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? (JEE Main 2014)
(a) A metre scale.
(b) A Vernier calliper where 10 divisions of Vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 mm
(c) A screw gauge having 100 divisions in the circular scale and pitch of 1 mm.
(d) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.
Q16. The period of oscillation of a simple pendulum is
Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in determination of g is (JEE Main 2015)
(a) 1%
(b) 5%
(c) 2%
(d) 3%
Q17. A student measures the time period of 100 oscillations of a simple pendulum 4 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 reported mean time should be (JEE Main 2016)
(a) 92 ± 1.5 s
(b) 92 ± 3 s
(c) 92 ± 2 s
(d) 92 ± 5.0 s
Q18. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought together in contact, 5th division coincides with the main scale line and zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and 25th division coincides with the main scale line? (JEE Main 2016)
(a) 0.70 mm
(b) 0.50 mm
(c) 0.75 mm
(d) 0.80 mm
Solutions
Solution 1:
| Sr. No. | Quantity | Dimensions |
| 1 | Torque | [M1L2T-2] |
| 2 | Work | [M1L2T-2] |
| 3 | Stress | [M1L-1T-2] |
| 4 | Energy | [M1L2T-2] |
| 5 | Force | [M1L1T-2] |
From table we can see that option (a) is correct. option
Do you want to know how the dimensions of above physical quantities are derived? Click Here
Solution 2:
Dimensions of 1/μoεo are [L2T-2]. Thus option (c) is correct option.
Solution 3:
| Sr. No. | Quantity | Dimensions |
| 1 | Torque | [ML2T-2] |
| 2 | Work | [ML2T-2] |
| 3 | Momentum | [MLT-1] |
| 4 | Planck’s constant | [ML2T-1] |
| 5 | Stress | [ML-1T-2] |
| 6 | Young’s modulus | [ML-1T-2] |
| 7 | Speed | [L1T-1] |
| 8 | (μoεo)-1/2 | [L1T-1] |
From table we can see that option (c) is correct option.
Do you want to know how the dimensions of above physical quantities are derived? Click Here
Solution 4:
By Newton’s law of viscosity, the viscous force (drag) is given by
Hence the correct option is (a)
Solution 5:
| Sr. No. | Quantity | Dimensions |
| 1 | Impulse | [MLT-1] |
| 2 | Momentum | [MLT-1] |
| 3 | Angular Momentum | [ML2T-1] |
| 4 | Planck’s constant | [ML2T-1] |
| 5 | Work | [ML2T-2] |
| 6 | Torque | [ML2T-2] |
| 7 | Moment of Inertia | [ML2] |
| 8 | Moment of force | [ML2T-2] |
From table we can see that option (d) is correct option.
Solution 6:
We have,
Force (F) = Charge (Q) × Magnetic Field (B) × Velocity (v)
∴ B = F/Qv = F Q-1 v-1
∴ [B] = [MLT-2] [C]-1[LT-1]-1
∴ [B] = [MLT-2] [C-1] [L-1T1]
∴ [B] = [MT-1 C-1]
Hence (c) is the correct option
Solution 7:
Momentum = Mass x Velocity
∴ Momentum = 3.513 kg x 5.00 ms-1
∴ Momentum = 17.565 kg ms-1
∴ Momentum = 17.6 kg ms-1 (Corrected to 3 significant figures)
Hence (a) is correct option
Solution 8:
Pitch of screw = 1mm/2 turns = 0.5 mm
Least Count of screw gauge = pitch/Total divisions on circular scale
∴ Least Count (L.C.) = 0.5/50 = 0.01 mm
Now, Reading = Main Scale Reading + Circular Scale Reading x Least Count – Zero error
∴ Reading = 3 + 35 x 0.01 – (-0.03)
∴ Reading = 3 + 0.35 + 0.03 = 3.38 mm)
Hence (d) is correct option
Solution 9:
Given, 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale
1 V.S.D. = 29/30 M.S.D.
∴ Least Count = 1 M.S.D. – 1 V.S.D.
∴ Least Count = 1 M.S.D. – 29/30 M.S.D.
∴ Least Count = 1/30 M.S.D. = 1/30 x 0.5
∴ Least Count =1/60 degree = 1 minute
Hence (d) is correct option
Solution 10:
Number of significant figures in 23.023 are 5
Number of significant figures in 0.0003 are 1
Number of significant figures in 2.1 x 10-3 are 2
Hence (a) is correct option
Solution 11:
Least count of screw gauge = 1/100 = 0.01 mm
Diameter of wire = M.S.R. + L.C. x C.S.R.
∴ Diameter of wire = 0 + 0.01 x 52 = 0.52 mm = 0.052 cm
Hence (a) is correct option
Solution 12:
By Ohm’s law, V = IR
∴ R = V/I
∴ % error in R = % error in V + % error in I
∴ % error in R = 3% + 3% = 6%
Hence (a) is correct option
Solution 13:
Total divisions on the Vernier scale is 30 and match with 30 divisions of the main scale.
1 V.S.D. = 29/30 M.S.D.
∴ Least Count = 1 M.S.D. – 1 V.S.D.
∴ Least Count = 1 M.S.D. – 29/30 M.S.D.
∴ Least Count = 1/30 M.S.D. = 1/30 x 0.5
∴ Least Count =1/60 degree
Angle of prism = M.S.R. + L.C. x V.S.R.
∴ Angle of prism = 58.5 + (1/60) x 9
∴ Angle of prism = 58.5 + 0.15 = 58.65 degree.
Hence (c) is correct option
Solution 14:
By Coulomb’s law
Hence dimensions of electrical permittivity are [M-1L-3T4A²]
Hence (b) is correct option
Solution 15:
Consider option (a), themetre scale:
The least count of metre scale is 1 mm = 0.1 cm.
∴ This is not the instrument as it gives reading upto one decimal
Consider option (b), the Vernier calliper:
Least count of main scale = 1/10 = 0.1 cm
1 V.S.D. = 9/10 M.S.D.
∴ Least Count of Vernier calliper = 1 M.S.D. – 1 V.S.D.
∴ Least Count of Vernier calliper = 1 M.S.D. – 9/10 M.S.D.
∴ Least Count of Vernier calliper = 1/10 M.S.D. = (1/10) x 0.1 cm = 0.01 cm. This is one of the correct answer.
Consider option (c), the screw gauge:
Least Count of this screw gauge = 1/100 = 0.01 mm = 0.001 cm
∴ This is not the instrument as it gives reading upto three decimals
Consider option (d) screw gauge:
Least Count of this screw gauge = 1/50 = 0.02 mm = 0.002 cm
∴ This is not the instrument as it gives reading upto three decimals
Hence (b) is correct option
Solution 16:
g = 4π2L/T2
(Δg/g) x 100 = (ΔL/L) x 100 + 2(ΔT/T) x 100
(Δg/g) x 100 = (0.1 cm/20.0 cm) x 100 + 2(1 s/90 s) x 100
(Δg/g) x 100 = 2.72% approx. 3%
Hence (d) is a correct option
Solution 17:
Mean reading = (90 + 91 + 95 + 92)/4 = 368/4 = 92 s
Mean error = ΔT = (|ΔT1| + |ΔT2| + |ΔT3| + |ΔT4|)/4
ΔT = (|90 – 92| + |91 – 92| + |95 – 92| + |92 – 92|)/4
ΔT = (|-2| + |-1| + |3| + |0|)/4
ΔT = (2 + 1 + 3 + 0)/4 = 6/4 = 1.5
Thus reported mean time = 92 ± 1.5 s
Hence (a) is the correct option.
Solution 18:
Least Count = pitch/no. of divisions on circular scale
Least Count = 0.5/50 = 0.01 mm
Zero Error = – 0.01 x 5 = – 0.05 mm
Thickness = M.S.R. + L.C. x C.S.R. – Zero error
Thickness = 0.5 + 0.01 x 25 + 0.05
Thickness = 0.5 + 0. 25 + 0.05 = 0.80 mm
Hence (d) is the correct option.