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 Multiple Choice QuestionsMultiple Choice Questions

1.

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 the main scale corresponds 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

  • 58.59 degree

  • 58.77 degree

  • 58.65 degree

  • 58.65 degree


C.

58.65 degree

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2.

Which one of the following represents the correct dimensions of the coefficient of viscosity?

  • ML−1 T−2

  • MLT−1

  • ML−1 T−1

  • ML−1 T−1


C.

ML−1 T−1

straight eta space equals space fraction numerator straight F over denominator straight A left parenthesis increment straight V subscript straight x divided by increment straight z right parenthesis end fraction
therefore space Dimensions space of space straight eta
space equals space fraction numerator Dimensions space of space force over denominator Dimensions space of space area space straight x space Dimensions space of space velocity space gradient end fraction
space equals space fraction numerator left square bracket MLT to the power of negative 2 end exponent right square bracket over denominator left square bracket straight L squared right square bracket left square bracket straight T to the power of negative 1 end exponent right square bracket end fraction space equals space left square bracket ML to the power of negative 1 end exponent straight T to the power of negative 1 end exponent right square bracket
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3.

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then

  • 0] = [M-1L-3T2A]

  • 0] = [M-1L-3T4A2]

  • 0] =[M-2L2T-1A-2]

  • 0] =[M-2L2T-1A-2]


B.

0] = [M-1L-3T4A2]

From Coulomb's Law, F

straight F space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight R squared end fraction
straight epsilon subscript straight o space equals space fraction numerator straight q subscript 1 straight q subscript 2 over denominator 4 πFR squared end fraction

On Substituting the units, we get


straight epsilon subscript straight o space equals space fraction numerator straight C squared over denominator straight N minus straight m end fraction space equals space fraction numerator left square bracket AT right square bracket squared over denominator left square bracket MLT to the power of negative 2 end exponent right square bracket left square bracket straight L squared right square bracket end fraction space left parenthesis 4 straight pi space is space dimensionless right parenthesis
space equals space left square bracket straight M to the power of negative 1 end exponent straight L to the power of negative 3 end exponent straight T to the power of 4 straight A squared right square bracket

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4.

The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are

  • 5,1,2

  • 5,1,5

  • 5,5,2

  • 5,5,2


A.

5,1,2

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5.

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm
Circular scale reading: 52 division
Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is

  • 0.052 cm

  • 0.026 cm

  • 0.005

  • 0.005


A.

0.052 cm

Least count of screw gauge =1/100 mm = 0.01 mn
Diameter - Divisions on cirular scale × least count + main scale reading = 52 × 1/100 + 0
100 = 0.52 mm
diameter = 0.052 cm

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6.

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 scale as 35. The diameter of the wire is

  • 3.32 mm 

  • 3.73 mm

  • 3.67 mm

  • 3.67 mm


D.

3.67 mm

Diameter = M.S.R. + C.S.R × L.C. + Z.E.
= 3 + 35 × (0.5/50) + 0.03
= 3.38 mm

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7.

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

  • A meter scale.

  • A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

  • A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

  • A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.


B.

A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

If student measures 3.50 cm, it means that there is an uncertainty of order 0.01 cm
For vernier scale with 1 MSD = 1mm
and 9 MSD = 10 VSD
LC of vernier scale with 1 MSD - 1 VSD
1 over 10 open parentheses 1 minus 9 over 10 close parentheses
space equals space 1 over 100 cm

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8.

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 scale as 35. The diameter of the wire is 

  • 3.32 mm

  • 3.73 mm

  • 3.67 mm

  • 3.67 mm


D.

3.67 mm

Diameter = M.S.R. + C.S.R × L.C. + Z.E.
= 3 + 35 × (0.5/50) + 0.03 = 3.38 mm 

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9.

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

  • one minute

  • half minute

  • One degree

  • One degree


A.

one minute

1VSD = 29/30 MSD
L.C. = 1 MSD – 1 VSD
= 1/30 MSD
1 over 30 space straight x 0.5 space equals space 1 over 60 space space equals space one space minute

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10.

The dimension of magnetic field in M, L, T and C (Coulomb) is given as

  • MLT−1C−1

  • MT2C−2

  • MT−1C−1

  • MT−1C−1


C.

MT−1C−1

F = qvB
B = F/qv
= MC−1 T−1

176 Views