JEE Main 2016 (Offline) | Units & Measurements Question 153 | Physics

A student measures the time period of 100 oscillations 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 mean time should be :

92 $$ \pm $$ 1.8 s

92 $$ \pm $$ 3 s

92 $$ \pm $$ 2 s

92 $$ \pm $$ 5.0 s

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)

-1

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 in contact, the 45^th division coincides with the main scale line and that the 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 the 25^th division coincides with the main scale line?

0.70 mm

0.50 mm

0.75 mm

0.80 mm

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-1

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

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

A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

A meter scale.

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

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)

-1

Let [$${\varepsilon _0}$$] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

$${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]$$

$${\varepsilon _0} = $$$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$

$${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]$$

$${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]$$

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