Current Electricity Quiz 13

Current Electricity Quiz-13

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. The resistor of resistance R is connected to 25 V supply and heat produced in it is 25 Js^-1. The value of R is?

•  225 Ω
•  1 Ω
•  25 Ω
•  50 Ω

Solution
H= V²/R ⇒R=V²/H = (25)²/25=25 Ω

Q2. A galvanometer can be converted into a voltmeter by connecting?

•  Low residence in parallel
•  Low residence in series
•  High residence in parallel
•  High residence in series

Solution
To convert a moving coil galvanometer into a voltmeter, a high resistance is connected in series with it.

Q3. An electric heater of 1.08 Kw is immersed in water. After the water has reached a temperature of 100℃, how much time will be required to produce 100 g of steam?

•  420 s
•  210 s
•  105 s
•  50 s

Solution
Heat produced by heater per second =1.08×10³J Heat taken by water to form steam mL =100×540 cal =100×540×4.2 J ∴1.08×10³×t=100×540×4.2 or t=(100×540×4.2)/(1.08×10³ ) = 210 s

Q4. Two voltameters, one of copper and another of silver, are joined in paralleled. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are z₁ and z₂ respectively, the charge which flows through the silver voltameter is?

•  q/(1+z₁/z₂ )
•  q/(1+z₂/z₁ )
•  q z₁/z₂
•  q z₂/z₁

Solution
We know m=zq ⇒z∝1/q ∴ z₂/z₁ =q₁/q₂ Total charge q=q₁+q₂ q/q₂ =q₁/q₂ +1 ⇒q₂=q/((1+q₁/q₂ ) ) ∴q₂=q/((1+z₂/z₁ ) )

Q5. A primary cell has an e.m.f. of 1.5 volt, when short-circuited it gives a current of 3 ampere. The internal resistance of the cell is?

•  4.5 ohm
•  2 ohm
•  0.5 ohm
•  1/4.5 ohm

Solution
 Short circuit current i_SC=E/r⇒3=1.5/r⇒r=0.5Ω

Q6. An immersion heater is rated 418 W. It should heat a litre of water from 10℃ to 30℃ in nearly?

•  44 s
•  100 s
•  200 s
•  400 s

Solution
418t=1×4180×20 or t=200 s Note that in this case, C=418 J/kg^-1) [℃^-1)] J=4.18 J cal^-1

Q7. The figure shows a network of currents. The magnitude of currents is shown here. The current i will be?

•  3 A
•  13 A
•  23 A
•  -3 A

Solution
By Kirchhoff’s current law

Q8. A block has dimensions 1 cm,2 cm,3 cm. Ratio of the maximum resistance to minimum resistance between any point of opposite faces of this block is?

•  9 :1
•  1 :9
•  18 :1
•  1 :6

Solution
a=1,b=2,c=3 ⇒R_max=(ρ.L)/A=(ρ.c)/(a.b)

R_min=(ρ.L'')/A''=(ρ.a)/(b.c) ⇒R_max/R_min =((ρ.c)/(a.b))/((ρ.a)/(b.c))=c/a×c/a⇒c²/a² =(c/a)²=(3/1)²=9/1

Q9. In the given figure. A,B and C are three identical bulbs. When the switch S is closed?

•  The brightness of bulb A does not change and that of B decreases
•  The brightness of bulb A increases and that of B decreases
•  The brightness of A increases bulb B does not glow
•  The brightness of both bulbs A not B decrease

Solution
When switch Sis pressed, the resistance of circuit decreases. Hence, the current in bulb A will increase but the current in bulb B will decrease. Hence, the brightness of bulb A will increase and that of bulb B will decrease.

Q10. The length of a potentiometer wire is 5m. An electron in this wire experiences a force of 4.8×10^-19N, emf of the main cell used in potentiometer is?

•  3 V
•  15 V
•  1.5 V
•  5 V

Solution
Force =Electric intensity ×charge =(Potential diffsrence)/distance×charge ∴ 4.8×10^-19=V/5×1.6×10^-19 or V=15 volt