Question

In a Roman steelyard, the weight of the Rider is 10 N and the scale is calibrated in mm. If the load is suspended 8 cm away from the point of suspension, then the least count of the instrument is _________. (Take g = 10 m s^-2)

• A. 4 N
• B. $\frac{1}{8}N$
• C. 8 N
• D. $\frac{1}{4}N$

Answer 1

The correct option is B

$\frac{1}{8}N$

Step 1, Given data

Weight of the rider (R) 10 N

Scale is calibrated in mm so $Callibration=1mm$

Load is suspended at $8cm=80mm$

Step 2, Finding the least count

We know,

$LC=\frac{R\times callibration}{Dis\tan ceoftheload}$

Putting all the values

$LC=\frac{10\times 1}{80}=\frac{1}{8}$N

Therefore the least count of the instrument is $\frac{1}{8}N$

Hence the correct option is (B) $\frac{1}{8}N$