Physics, asked by kvnewcantt33, 8 months ago

A person decides to use his bath tub water to generate
electric power to run a 40 W bulb. The bath tub is
located at a height of 10 m from the ground and it
holds 200 litres of water. He installs a water driven
wheel generator on the ground. The minimum rate at
which the water should drain from the bath tub to just
light the bulb is [Efficiency of generator is 90% and take
g = 9.8 ms^-2]​

Answers

Answered by sanupeharkar
0

Answer:

To power a 40 watt bulb, the generator should provide 40 J of energy per second.

Now if the generator efficiency is 90% then the input energy per second to generator should be-

E

in

=

0.9

E

out

or,

E

in

=

0.9

40

J or,

E

in

=

9

400

J

Now assume m

r

mass of water fall on wheel per second, then energy lost by water or energy transferred to wheel per second is

E

transferred

=m

r

×g×h

and we have

E

in

=E

transferred

thus

E

in

=m

r

×10×10 or,

9

400

=m

r

×10×10 or,

m

r

=

9

4

Kg

Thus the mass fall rate is

9

4

Kg/s.

Now, density of water is 1 Kg/L thus 200 L of water has mass

m=200 Kg

Now total time that 200 L of water power the 40 W bulb is

t=

m

r

m

or,

t=

4/9

200

s or,

t=

4

200×9

s or,

t=450 s

thus total time the bulb can be kept on is 450 s.

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