Question

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
A. A metre scaleB. 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 cmC. A screw gauge having 100 divisions in the circular scale and pitch as 1 mmD. A screw guage having 50 divisions in the circular scale and pitch as 1 mm

Answer 1

Final Answer : B

Logic Used :
1) Since, Student Measured
the Length of Rod : 3.50 cm
=>Least Count of Instrument :0.01cm

Steps:
Now, we will find least count in all four -options :
A: Metre -scale :
Least Count ,LC : 0.1cm
=>Not used in experiment :

B: Vernier Caliper :
LC of Main scale =0.1cm

10VSD =9 MSD
Least Count, LC = 1 MSD - 1 VSD
= 1 MSD - 9/10 MSD
= 1/10MSD =0.1*0.1=0.01 CM
Hence, This matches with that of student measurement .√√

C: Screw Gauge :
Pitch = 1 mm
No. of circular divisions : 100
So, Least count = 1/100 mm =0.01mm
=> Not used

D: Screw Gauge :
Pitch : 1 mm
No. of circular divisions : 50
So, Least Count : 1/50 mm :0.02mm
=> Not used

Hence, Option B is right

Answer 2

Answer:

The right answer will be Option (b)