[2 Mark Question Answer]

Question 512 Marks

A pump is used to lift $500 kg$ of water from a depth of $80 m$ in $10 s$. Calculate: The power rating of the pump if its efficiency is $40 \%$. (Take
$
\begin{aligned}
& g=10 ms 2) \\
& {\left[\text
{ Hint : Efficiency }=\frac{\text { useful power }}{\text { power input }}\right]}
\end{aligned}
$

Answer

$\begin{aligned} & \text { Efficiency }=\frac{\text { useful power }}{\text { power input }} \\ & \text { Efficiency }=40 \%=0.4 \\ & 0.4=\frac{40 kW }{\text { Power input }} \\ & \text { Power input }=\frac{40 kW }{0.4}=100 kW \end{aligned}$

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Question 522 Marks

A pump is used to lift 500kg of water from a depth of 80m in 10s. Calculate:
The power at which the pump works

Answer

Power at which pump works =
$ \frac{\text { work done }}{\text { Time taken }}=\frac{ mgh }{ t }=\frac{50 \times 10 \times 80 J }{10 s }=\frac{4 \times 10^5}{10}=40 kW $

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Question 532 Marks

QUESIION A water pump raises 50 litres of water through a height of 25 m in 5 s . Calculate the power of the pump required. (Take $g =10 N kg ^{-1}$ and density of water $=1000 kg m ^{-3}$ )

Answer

$ d=\frac{m}{v}$
$m=v \times d$
$m=50 L \times 1000 kg / m ^{-3}$
$m=50 m ^3 \times 1000 kg / m ^{-3}$
$m=50 kg$
Work done in raising $50 kg$ water to a height of $25 m$ against the force of gravity is:
$\text { Power }=\frac{\text { work done }}{\text { time }}=\frac{ mgh }{ t }$
$P=\frac{50 \times 10 \times 25}{5}$
$=2500 W$

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Question 542 Marks

An electric heater of power 3 KW is used for 10 h. How much energy does it consume? Express your answer in (i) kWh, (ii) joule.

Answer

Energy consumed = power $x$ time
1) Energy $=3 kW \times 10 h =30 kWh$
2) 1 kilowatt hour $( kWh )=3.6 \times 10^6 J$
$
\begin{aligned}
& 30 kWh =30 \times 3.6 \times 10^6 J \\
& =1.08 \times 10^8 J
\end{aligned}
$

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Question 552 Marks

A machine raises a load of 750 N through a height of 16 m in 5 s. calculate:
(i) energy spent by machine,
(ii) power at which the machine works.

Answer

(i) Energy spent by machine or work done = F S
Work, $W=750 \times 16=12000 J$
(ii) Power spent $=\frac{\text { work done }}{\text { time taken }}=\frac{12000 j }{5 s }=2400 w$

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Question 562 Marks

A weight lifter a load of 200 kgf to a height of 2.5 m in 5 s. Calculate: (i) the work done, and (ii) the power developed by him. Take g = 10 N kg-1

Answer

Force= mg = 200 × 10 = 2000 N
Distance, S = 2.5m
Time , t = 5 s
(i) Work done, W= F S
W = 2000 × 2.5m = 5000J
Power developed = work donetime taken= 5000J5s=1000W

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Question 572 Marks

A man spends $7.4\ kJ$ energy in displacing a body by $74\ m$ in the direction in which he applies force, in $2.5$ s. Calculate
the power spent (in H.P) by the man.

Answer

$\text { Power spent }=\frac{\text { work done }}{\text { time taken }} $
$ P=\frac{7.4 \times 10^3}{2.5} $
$ P=2960 W $
$ \therefore 1 H . P =746 W$
$ \therefore 1 W =\frac{1}{746} HP $
$ \therefore 2960 W =\frac{2960}{746} HP$
$\therefore 2960 W =3.97 HP$

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Question 582 Marks

A man spends 6.4 KJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s Calculate:
the force applied

Answer

Work done by man $=7.4 kJ$
Distance moved, $S =74 m$
Work done by the man= Force $x$ distance moved in direction of force
Work, $W = F \times S$
$7.4 \times 10^3= F \times 74$
$F =\frac{7.4 \times 10^3}{74}$
$F =100 N$

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Question 592 Marks

A man raises a box of mass 50kg to a height of 2m in 20s, while another man raises the same box to the same height in 50s. Calculated:
(i) the work done, and
(ii) the power developed by each man. Take $g = 10N kg^{-1}.$

Answer

(i) Work done $=50 \times 10 \times 2=1000 J$
(ii)
Power developed by first $\operatorname{man}=\frac{1000}{20}=50 W$
Power developed by second man $=\frac{1000}{50}=20 W$

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Question 602 Marks

A man raises a box of mass 50kg to a height of 2m in 20s, while another man raises the same box to the same height in 50s. Compare:
(i) the work done, and
(ii) the power developed by them.

Answer

(i) Work done to raise the block of mass 50kg is same for both.

(ii) Power= Work done/time. As the time taken by the first man is less therefore power developed is more.

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Question 612 Marks

It takes 20 s for a person A of mass 50 kg to climb up the stairs, while another person B does the same in 15 s. Compare the Work done.

Answer

It takes $20 s$ for a person A to climb up the stairs, while another person B does the same in $15 s$. Compare the (i) Work done and (ii) power developed by the persons $A$ and $B$.
$\frac{\text { work done by } A }{\text { work done by } B }=\frac{1}{1}=1: 1$

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Question 622 Marks

A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute, Calculate:
(i) the work done and
(ii) power spent.

Answer

Total distance covered in 30 steps, $S =30 \times 20 cm =600 cm =6 m$ Work done by the boy in climbing= Force $x$ distance moved in direction of force
Work, W= F x S $=350 \times 6=2100 J$
Power developed $=\frac{\text { work done }}{\text { time taken }}=\frac{2100 j }{60 s } 35 w$

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Question 632 Marks

Rajan exerts a force of $150$ N in pulling a cart at a constant speed of $10$ m/s. Calculate the power exerted.

Answer

Given that
Force $= 150 N$
Velocity $= 10 m s^{-1}$
$\therefore$ Power $= F x v$
$= 150 N x 10ms^{-1}$
$= 1500 W$

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Physics [2 Mark Question Answer]