Question

The linear mass density of a ladder of length L increases uniformly from one end A to other end B, form an expression for linear mass density as a function of distance $x$ from end A where linear mass density ${{\lambda }_{0}}$. The density at one end being twice that of the other end.
A. ${{\lambda }_{\left( x \right)}}={{\lambda }_{0}}+\dfrac{\lambda x}{2L}$
B. ${{\lambda }_{\left( x \right)}}={{\lambda }_{0}}+\dfrac{\lambda x}{L}$
C. ${{\lambda }_{\left( x \right)}}=2{{\lambda }_{0}}+\dfrac{\lambda x}{L}$
D. ${{\lambda }_{\left( x \right)}}={{\lambda }_{0}}+\dfrac{2\lambda x}{L}$

Answer 1

Hint: First of all. we will consider any linear equation having constants A and B, and then we will consider a point at distance x where the mass density of the ladder will become zero. Then we will find the exact distance of that point from point A and find the answer.

Complete Step-by-Step solution:
 In the question we are given that the linear mass density of a ladder of length L increases uniformly from one end A to other end B and we are asked to find the linear mass density at distance x from the point A.
First of all, we will consider a ladder of length l, having two points A and B as shown in the figure,

Now, the linear equation which can be assumed for mass density distribution can be given as,
$\lambda =Ax+B$………………………..(i)
Where A and B are constants, x is distance from end A and $\lambda $ is linear mass density.
Now, for ${{\lambda }_{A}}$i.e. linear mass density at point A, the value of distance x is 0 as it is the origin so, expression can be given as,
${{\lambda }_{A}}=A\left( 0 \right)+B$
$\Rightarrow {{\lambda }_{A}}=B$……………(ii)
Now, in question it is given that value of linear mass density at point A is ${{\lambda }_{0}}$so, substituting this value in expression (ii) we will get,
$B={{\lambda }_{0}}$ …………..(iii)
Now, for point B i.e. linear mass density at point B, the value of distance x will be L as it is at distance L from point A and it can be seen mathematically as,
${{\lambda }_{B}}=A\left( L \right)+B$ …………………(iv)
Now, in question it is given that the value of linear mass density at point B is tice the value of linear mass density at point A, so substituting this value in expression (iii) we will get,
$2{{\lambda }_{0}}=A\left( L \right)+B$
Now, on substituting the value of B from expression (iii) we will get,
$2{{\lambda }_{0}}=A\left( L \right)+{{\lambda }_{0}}$
$\Rightarrow 2{{\lambda }_{0}}-{{\lambda }_{0}}=A\left( L \right)$
\[\Rightarrow A\left( L \right)={{\lambda }_{0}}\Rightarrow A=\dfrac{{{\lambda }_{0}}}{L}\] ………….(v)
Now, substituting the derived values of A and B in expression (i) we will get,
$\lambda =\dfrac{{{\lambda }_{0}}}{L}x+{{\lambda }_{0}}$
Hence, the value of linear mass density at distance x from point A can be written as, ${{\lambda }_{x}}={{\lambda }_{0}}+\dfrac{{{\lambda }_{0}}x}{L}$.
Thus, option (b) is correct.

Note: In such a type of question students might make mistakes in taking the point at distance x from side A or B and they might take the distance x from B instead of A and due to the answer may change completely so students must read the question carefully and solve accordingly.