ICSE Class 10 Physics 2017 Solved Paper

A pulley system with $V R=4$ is used to lift a load of $175 {\kgf}$ through a vertical height of $15 {\m}$ . The effort required is $50 {\kgf}$ in the downward direction. $(g=10 {\N} {\kg}^{-1})$
Calculate:

Question : 28 of 65

Marks: +1, -0

Distance moved by the effort.

Solution:

Given : V.R.

=4

, Load

=175 {\kgf}

, displacement of load

(d_{ {\L}}) =15 {\m}

., Effort

=50

kgf,

g=10 {\Nkg}^{-1}

Distance moved by the effort

(d_{ {\E}}) =

?
We know,

\;\text " V.R. "\;\;=\;{d_{ {\E}}}/{d_{ {\L}}}

⇒ \;\; {\d}_{ {\E}}\;=\;\text " V.R. "\; × d_{ {\L}}=4 × 15

\;=60 {\m}

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