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A metre scale is balanced on a knife edge at its centre. When two coins, each of mass \mathrm{10\, g} are put one on the top of the other at the \mathrm{10.0\, cm} mark the scale is found to be balanced at \mathrm{40.0\, cm} mark. The mass of the metre scale is found to be \mathrm{x\times 10^{-2}kg}. The value of \mathrm{x} is __________.

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer



Let the mass of metre scale &
Coin be M and m respectively.

For scale to be balanced ,
\mathrm{\Sigma F= 0\; and\; \Sigma \tau = 0}

R is the reaction force on a knife edge
\mathrm{\Sigma F= 0}
\mathrm{R= 2mg+Mg}
\mathrm{\Sigma \tau = 0}

about knife edge
\mathrm{\left ( 2\, mg \right )\left ( 30 \right )\left ( \leftarrow \right )+Mg\times 10\left ( \rightarrow \right )= 0}
\mathrm{m= \frac{M}{6}\Rightarrow M= 6M}
   \mathrm{= 609}
\mathrm{M= 6\times 10^{-2}kg}

\mathrm{\therefore x= 6}
 

Posted by

Ramraj Saini

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