Q

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

• Option 1)

10.04 Volt

• Option 2)

Zero Volt

• Option 3)

2.9 Volt

• Option 4)

13.3 Volt

S solutionqc

As discussed in

Power dissipiated in external resistance -

$P=(\frac{E}{R+r})^{2}R$

-

$P=\frac{v^{2}}{R}\:=R\:=\frac{v^{2}}{P}=\frac{120\times 120}{60}\:=240\Omega$

Power of heater = $240\Omega$

$R=\frac{120\times 120}{240}=60\Omega$

Voltage across bulb before heater switched on

$v_{1}=\frac{240}{246}\times 120 = 117.73v$

When Heater and bulb in parallel then $R_{eq}=\frac{240*60}{300}=\frac{240}{5}$

Voltage across bulb after heater switched on:

$v_{2}=I_{net}R_{eq}=(\frac{120}{6+(\frac{240}{5})})\frac{240}{5}=(\frac{120*5}{270})\frac{240}{5}=\frac{320}{3}=106.66$

$v_{2}-v_{1}=117.073-106.66=10.04$

Option 1)

10.04 Volt

Option 2)

Zero Volt

Option 3)

2.9 Volt

Option 4)

13.3 Volt

74 Views

Find out the appropriate words to fill in the blank:

The drunken man teased women and was consequentially ___ by them

• Option 1)

Pacified

• Option 2)

Appeased

• Option 3)

Neglected

• Option 4)

Castigated

S solutionqc

His teasing them would have resulted in his being punished or rebuked.

Option 1)

Pacified

incorrect option

Option 2)

Appeased

incorrect option

Option 3)

Neglected

incorrect option

Option 4)

Castigated

correct option

137 Views

Find out the correct meaning of the following phrase

Bite off more than you can chew

• Option 1)

To take a huge and difficult task

• Option 2)

To eat more than you can digest

• Option 3)

To let the expenditure exceed the resources

• Option 4)

To work very hard

S solutionqc

The phrase means to take a huge and difficult task

Option 1)

To take a huge and difficult task

Correct option

Option 2)

To eat more than you can digest

incorrect option

Option 3)

To let the expenditure exceed the resources

incorrect option

Option 4)

To work very hard

incorrect option

109 Views

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

As we discussed in

If only conservative forces act on a system, total mechnical energy remains constant -

$K+U=E\left ( constant \right )$

$\Delta K+\Delta U=0$

$\Delta K=-\Delta U$

-

Total work done by the person in lifting the weigh = mgh

$=10 \times \9.8 \times \1 \times \1000$

$=\98 \times \10^{3}J$

Total mechanical energy produced by burning 1 kg fat  $=(3.8 \times10^{7})\times 0.20 = 7.6 \times 10^{6}J$

Total fat burn $=\frac{98 \times10^{3}}{7.6 \times 10^{6}}kg=12.89 \times 10^{-3} kg$

Option 1)

2.45×10−3 kg

This is an incorrect option.

Option 2)

6.45×10−3 kg

This is an incorrect option.

Option 3)

9.89×10−3 kg

This is an incorrect option.

Option 4)

12.89×10−3 kg

This is the correct option.

867 Views
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