Question

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, then the relative percentage error in the density is:
A) 0.9%.
B) 2.4%.
C) 3.1%.
D) 4.2%.

Answer 1

Screw gauge is an instrument which is used to measure the diameter of the wires it can also be used to find the measurement of the irregular lamina. The screw gauge consists of parameters like pitch and least count. The pitch is defined as the distance moved by the spindle per revolution when the head scale is moved. The least count is defined as the smallest measurement that a measuring device can measure.

Formula used: The screw gauge reading is given by,
$S \cdot G{\text{ Reading}} = M \cdot S{\text{ Reading + }}\left( {\dfrac{P}{{C \cdot S \cdot D}}} \right) \times C \cdot S \cdot D{\text{ Reading}}$
Where screw gauge reading is S.G, the main scale reading is M.S, the pitch is P, the C.S.D is circular scale division and the circular scale division reading is C.S.D.
The formula of the relative percentage error in the density is given by,
$\dfrac{{\Delta \rho }}{\rho } \times 100 = \left( {\dfrac{{\Delta M}}{M} + 3\dfrac{{\Delta D}}{D}} \right) \times 100$
Where change id density is $\Delta \rho$ the density is $\rho$ the change of mass is $\Delta M$ the mass of the body is $M$ the change of diameter is $\Delta D$ and the diameter is$D$.

Complete step by step answer:
It is given in the problem that the diameter of the ball is measured with a screw gauge, whose pitch is 0.5mm and there are 50 divisions on the circular scale the reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions if the measured mass of the ball has a relative error of 2%, then we need to find the relative percentage error in the density of the solid ball.
The pitch is given to be 0.5 mm, the circular scalar division is 50 and the main scale reading is 2.5 mm.
The formula of the least count is given by,
$L \cdot C = \dfrac{P}{{C \cdot S \cdot D}}$
Where least count is L.C the pitch is P and the circular scale division is C.S.D.
The least count of the screw gauge is equal to,
$\Rightarrow L \cdot C = \dfrac{P}{{C \cdot S \cdot D}}$
The pitch is equal to 0.5 mm and the circular scale division is equal to 50, so the least count will be,
$\Rightarrow L \cdot C = \dfrac{P}{{C \cdot S \cdot D}}$
$\Rightarrow L \cdot C = \dfrac{{0 \cdot 5}}{{50}}$
$\Rightarrow L \cdot C = 0 \cdot 01mm$………eq. (1)
The circular scale division reading is 20 and the relative error in percentage is 2%. The diameter of the ball will be equal to the screw gauge reading.
The screw gauge reading is given by,
$S \cdot G{\text{ Reading}} = M \cdot S{\text{ Reading + }}\left( {\dfrac{P}{{C \cdot S \cdot D}}} \right) \times C \cdot S \cdot D{\text{ Reading}}$
Where screw gauge reading is S.G, the main scale reading is M.S, the pitch is P, the C.S.D is circular scale division and the circular scale division reading is C.S.D.
The M.S reading is 2.5 mm, the pitch is 0.5mm, the circular scale division is 50 and the circular scale division reading is 20. Therefore the screw gauge reading is equal to,
$\Rightarrow S \cdot G{\text{ Reading}} = M \cdot S{\text{ Reading + }}\left( {\dfrac{P}{{C \cdot S \cdot D}}} \right) \times C \cdot S \cdot D{\text{ Reading}}$
$\Rightarrow S \cdot G{\text{ Reading}} = 2 \cdot 5{\text{ + }}\left( {\dfrac{{0 \cdot 5}}{{50}}} \right) \times 50$
$\Rightarrow S \cdot G{\text{ Reading}} = 2 \cdot 7mm$………eq. (2)
The density of the ball is given by,
$\rho = \dfrac{M}{V}$
Where density is $\rho$ the mass is $M$ and volume is$V$. Therefore the density is given by,
$\Rightarrow \rho = \dfrac{M}{V}$
$\Rightarrow \rho = \dfrac{M}{{\left[ {\left( {\dfrac{{4\pi }}{3}} \right) \cdot {{\left( {\dfrac{D}{2}} \right)}^3}} \right]}}$
The formula of the relative percentage error in the density is given by,
$\dfrac{{\Delta \rho }}{\rho } \times 100 = \left( {\dfrac{{\Delta M}}{M} + 3\dfrac{{\Delta D}}{D}} \right) \times 100$
Where change id density is $\Delta \rho$ the density is $\rho$ the change of mass is $\Delta M$ the mass of the body is $M$ the change of diameter is $\Delta D$ and the diameter is$D$.
The relative error of mass in percentage is 2 the change in diameter is least count from equation (1) which is 0.01mm and the value of the diameter will be 2.7mm from equation (2) then the percentage error of the density is given by,
$\Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = \left( {\dfrac{{\Delta M}}{M} + 3\dfrac{{\Delta D}}{D}} \right) \times 100$
$\Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = \left( {\dfrac{{\Delta M}}{M} \times 100 + 3\dfrac{{\Delta D}}{D} \times 100} \right)$
$\Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = \left( {2 + 3 \times \dfrac{{0 \cdot 01}}{{2 \cdot 7}} \times 100} \right)$
$\Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = 3 \cdot 1\%$.
The relative percentage error of the density is equal to$\dfrac{{\Delta \rho }}{\rho } \times 100 = 3 \cdot 1\%$. The correct answer for this problem is option C.

Note: It is advisable to students to understand and remember the formula of the screw gauge reading as it can be helpful for solving problems like these. The screw gauge can measure objects which cannot be measured by the vernier caliper and also with more precision. The precision of the screw gauge is 0.01 mm.