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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2008

MCQ (Single Correct Answer)
An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by
A
a vernier scale provided on the microscope
B
a standard laboratory scale
C
a meter scale provided on the microscope
D
a screw gauge provided on the microscope

Explanation

To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope
2

AIEEE 2008

MCQ (Single Correct Answer)
While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of $$18$$ $$cm$$ during winter. Repeating the same experiment during summer, she measures the column length to be $$x$$ $$cm$$ for the second resonance. Then
A
$$18 > x$$
B
$$x > 54$$
C
$$54 > x > 36$$
D
$$36 > x > 18$$

Explanation

For first resonant length $$v = {v \over {4{\ell _1}}} = {v \over {4 \times 18}}$$ (in winter)

For second resonant length

$$v' = {{3v'} \over {4{\ell _2}}} = {{3v'} \over {4x}}$$ (in summer)

$$\therefore$$ $${v \over {4 \times 18}} = {{3v'} \over {4 \times x}}$$

$$\therefore$$ $$x = 3 \times 18 \times {{v'} \over v}$$

$$\therefore$$ $$x = 54 \times {{v'} \over v}cm$$

$$v' > v$$ because velocity of light is greater in summer as compared to winter

$$\left( {v \propto \sqrt T } \right)$$

$$\therefore$$ $$x > 54\,cm$$
3

AIEEE 2008

MCQ (Single Correct Answer)
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
A
3.32 mm
B
3.73 mm
C
3.67 mm
D
3.38 mm

Explanation

Least count of screw gauge = $${{0.5} \over {50}}mm$$ = 0.01mm

Main scale reading = 3 mm

Vernier scale reading = 35

$$\therefore $$ Reading = [Main scale reading + circular scale reading $$\times$$ L.C] - (zero error)

= [3 + 35 $$\times$$ 0.01] - (-0.03) = 3.38 mm
4

AIEEE 2005

MCQ (Single Correct Answer)
In a potentiometer experiment the balancing with a cell is at length $$240$$ $$cm.$$ On shunting the cell with a resistance of $$2\Omega ,$$ the balancing length becomes $$120$$ $$cm$$. The internal resistance of the cell is
A
$$0.5\Omega $$
B
$$1\Omega $$
C
$$2\Omega $$
D
$$4\Omega $$

Explanation

The internal resistance of the cell,

$$r = \left( {{{{\ell _1} - {\ell _2}} \over {{\ell _2}}}} \right) \times R$$

$$ = {{240 - 120} \over {120}} \times 2 = 2\Omega $$

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