Question

Question :-

A block of mass m is pulled by a uniform chain of mass m tied to it by applying a Force F at the other end of the chain . The tension at a point which is at a distance of quarter of length of the chain from the free end , will be

a) $\texbf{\textsf{\dfrac{4F}{5}}}$
b) $\textbf{\textsf{\dfrac{7F}{8}}}$
c) $\textbf{\textsf{\dfrac{4F}{7}}}$
d) $\textbf{\textsf\dfrac{6F}{5}}}$

Diagram :-

$\setlength{\unitlength}{1mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(3,0){3cm}}\put(1,1){\boxed{\Huge\bf M}}}\put(7.2,2.2){\boxed{\qquad\qquad\;\;\;}}\put(26.6,2.48){\vector(3,0){0.5cm}}\put(33,2){\bf F}\put(22,4.8){\vector(3,0){4mm}}\put(22,4.8){\vector(-3,0){4mm}}\put(21,6.7){\sf ^l\!/_4 }\put(0,0){\line(1,-2){2mm}}\put(4,0){\line(1,-2){2mm}}\put(8,0){\line(1,-2){2mm}}\put(12,0){\line(1,-2){2mm}}\put(16,0){\line(1,-2){2mm}}\put(20,0){\line(1,-2){2mm}}\put(24,0){\line(1,-2){2mm}}\end{picture}$

Note :- Kindly see the question from web to see the diagram​

Answer 1

Explanation:

Mass of the block = m

Mass of the chain = m

Total mass = 2m

Force applied = F

$\rm As, \: Acceleration = \dfrac{Force}{Mass}$

$\rm Acceleration \: of \: the \: system \: (a) = \dfrac{F}{2m}$

Let the point at which tension is to be calculated is A.

Then,tension at point A will be due to mass of the block and the mass of the $\rm \dfrac{3}{4}$ of the chain.

As it is a uniform chain of mass " m " ,

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \rm Mass \: of \: chain\: \alpha \: length\: of \:the\: chain$

∴ Mass of $\rm\dfrac{3}{4}$ the length of the chain will be $\dfrac{3m}{4}$

Hence,the mass of the system on which the tension acts at point A will be $\rm m + \dfrac{3m}{4}$

Total mass = $\rm\dfrac{7m}{4}$

$\rm The\: acceleration \: of \: the \: whole \: system \: is \: \dfrac{F}{2m}$

∴The tension at point A = total mass at point A × acceleration at point A

Tension = $\rm \dfrac{7m}{4} \times \dfrac{F}{2m}$

$\:\rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \dfrac{7F}{8}$

∴ The tension at a point which is at a distance of quarter of length of the chain from the free end , will be $\rm \dfrac{7F}{8}$

Answer 2

Diagram:

$\setlength{\unitlength}{1mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(3,0){3cm}}\put(1,1){\boxed{\Huge\bf M}}}\put(7.2,2.2){\boxed{\qquad\qquad\;\;\;}}\put(26.6,2.48){\vector(3,0){0.5cm}}\put(33,2){\bf F}\put(22,4.8){\vector(3,0){4mm}}\put(22,4.8){\vector(-3,0){4mm}}\put(21,6.7){\sf ^l\!/_4 }\put(0,0){\line(1,-2){2mm}}\put(4,0){\line(1,-2){2mm}}\put(8,0){\line(1,-2){2mm}}\put(12,0){\line(1,-2){2mm}}\put(16,0){\line(1,-2){2mm}}\put(20,0){\line(1,-2){2mm}}\put(24,0){\line(1,-2){2mm}}\end{picture}$

Given:

• A block of mass m is pulled by a uniform chain of mass m tied to it by applying a Force F at the other end of the chain . The tension at a point which is at a distance of quarter of length of the chain from the free end.

Find:

• MCQ

Solution:

✒ Tension of the point is P = ?

✒ mass of the chain is M

✏️mass of L/4 chain = $\frac{M}{L} \times \frac{L}{4} = \frac{M}{4}$

✏ mass of rest chain =

$\sf M - \frac{M}{4} = \frac{4M - M }{4} \\ \sf \implies \frac{3M}{4}$

__________________

we, know

$\boxed{\sf F = ma}$

So,

$\to\sf F = ma$

$\to\sf F = M+M \times a$

$\to\sf a = \frac{f}{M+M} \\ = \sf \frac{f}{2M}$

__________________

Now,

$\boxed{\sf T = mass \times a}$

So,

$\to\sf T = (M + m) \times a$

$\to\sf T = (M + m) \times \frac{F}{2M}$

$\to\sf T = (M + \frac{3M}{4} ) \times \frac{F}{2M}$

$\to\sf T = (\frac{4M + 3M}{4} ) \times \frac{F}{2M}$

$\to\sf T = (\frac{7M}{4} ) \times \frac{F}{2M}$

$\to\sf T = \frac{7F}{8}$

Hence, answer is option b) 7F/8