An elevator accelerates upwards at a constant rate. A uniform string of length L and mass m supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance l from the ceiling is T. The acceleration of the elevator is :

Question

An elevator accelerates upwards at a constant rate. A uniform string of length L and mass m supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance l from the ceiling is T. The acceleration of the elevator is : (A) ( T / M + m -( m l) L )- g (B) ( T /2 M + m -( ml ) L )+ g (C) ( T / M +( ml ) L )- g (D) ( T /2 M - m +( ml ) L )- g

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6d8487b88d369f8f548970a2d5f399a3-.png" /> Let a be the acceleration of the lift. Mass of lower portion of string =( m / L )( L -l) ∴ T - Mg -( mg / L )( L -l)=( M +( m / L )( L -l)) a ∴ a =( T / M + m -( m l) L )- g

Q. An elevator accelerates upwards at a constant rate. A uniform string of length $L$ and mass $m$ supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance $l$ from the ceiling is $T$ . The acceleration of the elevator is :

$\frac{T}{M + m - \frac{m l}{L}} - g$

$\frac{T}{2 M + m - \frac{m l}{L}} + g$

$\frac{T}{M + \frac{m l}{L}} - g$

$\frac{T}{2 M - m + \frac{m l}{L}} - g$

Let $a$ be the acceleration of the lift.
Mass of lower portion of string
$= \frac{m}{L} (L - l)$
$∴ T - M g - \frac{m g}{L} (L - l) = (M + \frac{m}{L} (L - l)) a$
$∴ a = \frac{T}{M + m - \frac{m l}{L}} - g$

Q. An elevator accelerates upwards at a constant rate. A uniform string of length L and mass m supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance l from the ceiling is T. The acceleration of the elevator is :

M+m−Lml​T​−g

2M+m−Lml​T​+g

M+Lml​T​−g

2M−m+Lml​T​−g