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Standard VI

Mathematics

Multiplication Patterns

The linear ma...

Question

The linear mass density of a ladder of length $L$ increases uniformly from one end $A$ to other end $B$ . The distance of the centre of mass from end $A$ is $\frac{x}{y} L .$ Find $x + y$ .

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Solution

Let, $λ$ be the mass density=kx .
Here $, k$ is constant and $x$ is distance from $A$ .{since given that density varies linearly}
COM from $A = \frac{\sum (λ d x) x}{\sum λ d x} = \frac{2}{3} L$
By comparing with $\frac{x}{y} L$ ,
$x = 2, y = 3$
$x + y = 5$ .

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The linear mass density of a ladder of length L increases uniformly from one end A to other end B. The distance of the centre of mass from end A is xyL. Find x+y.

Let, λ be the mass density=kx .Here, k is constant and x is distance from A.{since given that density varies linearly}COM from A=∑(λdx)x∑λdx=23LBy comparing with xyL,x=2, y=3x+y=5.

The linear mass density of a ladder of length $L$ increases uniformly from one end $A$ to other end $B$ . The distance of the centre of mass from end $A$ is $\frac{x}{y} L .$ Find $x + y$ .

Let, $λ$ be the mass density=kx .
Here $, k$ is constant and $x$ is distance from $A$ .{since given that density varies linearly}
COM from $A = \frac{\sum (λ d x) x}{\sum λ d x} = \frac{2}{3} L$
By comparing with $\frac{x}{y} L$ ,
$x = 2, y = 3$
$x + y = 5$ .