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Question

A frictionless pulley of negligible weight is suspended from a spring balance. Masses of $$1kg$$ and $$5kg$$ are tied to the two ends of a string which passes over the pulley. The masses moves due to gravity. During motion, the reading of the spring balance will be
1169141_e8dc30b57c954eff9e5b3697dd730091.png


A
53kgwt
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B
103kgwt
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C
6kgwt
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D
3kgwt
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Solution

The correct option is B $$\dfrac{10}{3}kg\, wt$$
$$ \begin{array}{l} \text { The Reading of the } \\ \text { spring Balance is equal } \\ \text { to }(2 T) \text { where (T) } \\ \text { is the Tension in } \\ \text { the string. } \end{array} $$ 
$$ \text {* FBD of } 5 \mathrm{~kg} \text { Block } $$ 
$$ \begin{array}{l} \text { Let the Blocks move with } \\ \text { an acceleration a Downward. } \end{array} $$ $$ \Rightarrow \quad 50-T=5 a \quad-0 $$.
 $$ \text { * FBD of Ikg Block } $$ $$ \Rightarrow \quad T-10=1 \cdot a-(2) $$ 
$$ \begin{array}{l} \text { On Solving eq }-(1) \text { and }(2) \\ \text { we get } T=\frac{50}{3} N=\frac{5 g}{3} N \end{array} $$
 $$ \begin{array}{l} \Rightarrow 2 T=\frac{\log N}{3} N \\ \text { Hence Reading of Spring Balance in } k g=\frac{10}{3} k g \end{array} $$

2001356_1169141_ans_44c2154239774b2eb0d227334bc8903b.jpeg

Physics

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