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Three blocks are connected as shown on a horizontal frictionless surface , if $m_1=1kg \: \: , m_2= 8kg\: \: , m_3= 27kg\: \: and\: \: T_3=36N, \: \: T_2$ will be?

Option 1)
18N

Option 2)
9N

Option 3)
3.375N

Option 4)
1.75 N

Answers (3)

best_answer

As we learnt in

3 block connected with string -

- wherein

$a=\frac{F}{m_{1}+m_{2}+m_{3}}$

$T_{1}=\frac{(m_{2}+m_{3})F}{m_{1}+m_{2}+m_{3}}$

$T_{2}=\frac{m_{3}F}{m_{1}+m_{2}+m_{3}}$

$M_{1}=1kg$

$M_{2}=8kg$

$M_{3}=36N$

$T_{2}=?$

$T_{3}-T_{2}=36a$ ............(1)

$T_{2}-T_{1}=8a$ ..................(2)

$T_{1}=1a$ .................(3)

Adding equation (1),(2) & (3)

We get $T_{3}=36a$

$\therefore 36= 36a$

$a = 1ms^{-2}$

$T_{1}=1N$

Now, $T_{2}-T_{1}=8\times 1$

$T_{2}-1=8$

$\Rightarrow T_{2}=8+1$

$T_{2}=9N$

Option 1)

18N

Incorrect

Option 2)

9N

Correct

Option 3)

3.375N

Incorrect

Option 4)

1.75 N

Incorrect

Posted by

prateek

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If we draw the fbd of block m3 then we get the equation ;; T $\large ^{_{3}}$ - T $\large ^{_{2}}$ = M $\large ^{_{3}}$ *a ------------ EQ - 1

where (a) is acceleration of the entire system . Acceleration of the system ;; a = F/( M $\large ^{_{1}}$ + M $\large ^{_{2}}$ + M $\large ^{_{3}}$ )

= 36/36

a = 1 m/s $\large ^{2}$

From eq -----1 ;; 36 - T $\large ^{_{2}}$ = 27 * 1

T $\large ^{_{2}}$ = 9 Newtons.

Posted by

Thomson

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Answer 2

Posted by

Siddharth Kapoor

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