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JEE Mains Previous Years Questions with Solutions

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1

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
2

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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