Question

A wire of length 50 cm has a mass of 20 gm. It its radius is halved by stretching, its new mass per unit length will be
(a)$0.4g\text{ }c{{m}^{-1}}$
(b)$0.2kg\text{ }c{{m}^{-1}}$
(c)$0.1g\text{ }c{{m}^{-1}}$
(d)$0.2g\text{ }c{{m}^{-1}}$

Answer 1

Hint:In order to calculate the new mass per unit length, we need to know that on stretching, volume remains constant. Hence on decreasing the radius by half, the length of the wire becomes quadrupled.
It is by the formula that Mass, \[m=\rho V=\rho \left( \pi {{r}^{2}}l \right)~\],where \[\rho =\] density of material of wire, \[r=~\]radius of wire and \[l=~\]length of wire.
When its radius is halved by stretching, the new mass will be,
\[~m\prime =\rho \pi {{\left( \dfrac{r}{2} \right)}^{2}}l\prime \]
Thus,
\[\dfrac{{{m}'}}{m}=\dfrac{{{l}'}}{4l}\]

or \[\dfrac{{{m}'}}{l'}=\dfrac{m}{4l}\]
Hence, the new mass per unit length is
\[\dfrac{{{m}'}}{l'}=\dfrac{m}{4l}\]
\[=\dfrac{20}{4\times 50}=0.1g\text{ }c{{m}^{-1}}\]

Thus, answer to this question is option C

Additional Information: A fundamental unit is a unit adopted for measurement of a base quantity. A base quantity is one of a conventionally chosen subset of physical quantities, where no subset quantity can be expressed in terms of the others.Fundamental quantity: quantities which are independent of other physical quantities. ex: length, mass, time, current, amount of substance, luminous intensity, thermodynamic temperature, Derived quantity: quantities which depend on fundamental quantities.

Note:While solving this question, we should be careful with the values given in the question. It must be noted while solving that the formula used is from fundamental and derived units. we need to know that on stretching, volume remains constant. Hence on decreasing the radius by half, the length of the wire becomes quadrupled. The variable in the formula must be replaced with values given in the question.