IIT JEE Main 2016 Physics Question 22

IIT JEE Main 2016 Physics Question No 22

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 time sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th 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 25th division coincides with the main scale line?

(1) 0.70 mm (2) 0.50 mm (3) 0.75 mm (4) 0.80 mm

Solution

We know that,
Least Count (LC) = pitch/total divisions
Here, LC = 0.5/50
=> LC = 0.01 mm
Here the error is negative and is given by
error = 5 × 0.01 = 0.05
Reading = 0.5 + 25 × 0.01 + 0.05
= 0.5 + 0.25 + 0.05
= 0.80 mm

Answer

Correct answer to given problem No 22 of IIT JEE Main Physics paper of 2016 is option number (4) that is 0.80 mm.

IIT JEE Main 2016 Physics Question No 3

IIT JEE Main 2016 Physics Question No 3

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95 s and 92 s. If The minimum division in the measuring clock is 1s, then the reported mean time should be :

(1) 92 +/- 1.8 s
(2) 92 +/- 3 s
(3) 92 +/- 2 s
(4) 92 +/- 5.0 s

Solution

Mean value of observations i.e. time is obtained by,
Tmean = (90 + 91 + 95  + 92)/4
=> Tmean = 92s
Also absolute errors are as follows:
E1 = |92 – 90| = 2
E2 = |92 – 91| = 1
E3 = |92 – 95| = 3
E4 = |92 – 92| = 0
Therefore,  mean absolute error = (2 + 1 + 3 + 0)/4
= 1.5 s,  but since the least count of the clock is 1 second,  so it can’t count 1.5 s. Hence, it will count 2 seconds,  as an error.
(92 +/- 2) seconds is the correct option.

Correct answer to given problem No 3 of IIT JEE Main Physics paper of 2016 is option (3).