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ICSE Class 10 Physics Question 19 of 19

Machines — Question 10

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Question 10

A block and tackle system has the velocity ratio 3. Draw a labelled diagram of the system indicating the points of application and the directions of load L and effort E. A man can exert a pull of 200 kgf.

(a) What is the maximum load he can raise with this pulley system if its efficiency is 60%?

(b) If the effort end moves a distance 60 cm, what distance does the load move?

Answer

Given,

V.R = n (number of pulleys) = 3

Hence, number of pulleys = 3

Below is a labelled diagram of the system indicating the points of application and the directions of load L and effort E:

A block and tackle system has the velocity ratio 3. Draw a labelled diagram of the system indicating the points of application and the directions of load L and effort E. Machines, Concise Physics Class 10 Solutions.

(a) As we know,

η=1−wnE\eta = 1 - \dfrac{w}{nE}

Given,

Effort = 200 kgf

Efficiency = 60% = 0.6

Substituting the values in the formula for efficiency we get,

0.6=1−w3×200w600=1−0.6w=600×0.4⇒w=240 kgf0.6 = 1 - \dfrac{w}{3 \times 200} \\[0.5em] \dfrac{w}{600} = 1 - 0.6 \\[0.5em] w = 600 \times 0.4 \\[0.5em] \Rightarrow w = 240 \text{ kgf}

As we know,

Load=(n×E)−wLoad=(3×200)−wLoad=600−240⇒Load=360kgf\text{Load} = (n \times E) - w \\[0.5em] \text{Load} = (3 \times 200) - w \\[0.5em] \text{Load} = 600 - 240 \\[0.5em] \Rightarrow \text{Load} = 360kgf

(b) The velocity ratio of system is V.R = number of pulleys = 3

As we know,

V.R.=dEdLV.R. = \dfrac{d_E}{d_L} \\[0.5em]

Substituting the values in the formula for V.R. we get,

3=dEdLdL=dE3dL=603⇒dL=20cm3 = \dfrac{d_E}{d_L} \\[0.5em] d_L= \dfrac{d_E}{3} \\[0.5em] d_L= \dfrac{60}{3} \\[0.5em] \Rightarrow d_L= 20 cm \\[0.5em]