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ICSE Class 10 Physics Question 8 of 20

Solved 2019 Question Paper ICSE Class 10 Physics — Question 3

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Question 5(c)

The diagram below shows a pulley arrangement : [4]

The diagram below shows a pulley arrangement. Copy the diagram and mark the direction of tension on each strand of the string. What is the velocity ratio of the arrangement? If the tension acting on the string is T, then what is the relationship between T and effort E? If the free end of the string moves through a distance x, find the distance by which the load is raised. ICSE 2019 Physics Solved Question Paper.

(i) Copy the diagram and mark the direction of tension on each strand of the string.

(ii) What is the velocity ratio of the arrangement?

(iii) If the tension acting on the string is T, then what is the relationship between T and effort E?

(iv) If the free end of the string moves through a distance x, find the distance by which the load is raised.

Answer

(i) The direction of tension (T) is marked in the diagram:

The diagram below shows a pulley arrangement. Copy the diagram and mark the direction of tension on each strand of the string. What is the velocity ratio of the arrangement? If the tension acting on the string is T, then what is the relationship between T and effort E? If the free end of the string moves through a distance x, find the distance by which the load is raised. ICSE 2019 Physics Solved Question Paper.

(ii) V.R. = Distance moved by effortDistance moved by load\dfrac{\text{Distance moved by effort}}{\text{Distance moved by load}} = 2

(iii) Effort E balances the tensions therefore, T = E

(iv) From the relation,

V.R. = Distance moved by effortDistance moved by load\dfrac{\text{Distance moved by effort}}{\text{Distance moved by load}} = 2

If the free end of the string moves through a distance x then distance moved by load = ?

Substituting the values we get,

2=xDistance moved by load⇒Distance moved by load=x22 = \dfrac{\text{x}}{\text{Distance moved by load}} \\[0.5em] \Rightarrow \text{Distance moved by load} = \dfrac{x}{2}