Class 10 Science Chapter 11 MCQ

Explanation: Electric power is given by P = Voltage x Current SI unit of voltage = volt SI unit of current = Ampere So, its unit is volt ampere.

The SI unit of electric charge coulomb is equivalent to the charge contained in 6.25 x 10 to the power 18.

In series combination of resistances same current flows through each resistor where as in parallel combination same voltage exist across each resistor. Thus, the two resistors of resistances 2 Ω and 4 Ω are connected in series has same current flowing through it.

Here, all these statements are correct. Hence, your answer will be option [D].

Given resistors, R1 = 2 Ω and R2 = 4 Ω Voltage, V = 6 V1 Time, t = 5 s Resistance, R = 4 Ω Heat dissipated, H = ? Resistor, Rs = R1 + R2 = 2 + 4 = 6 Ω Current, I = V/R = 6/6 = 1A. Heat dissipated H = I¬^2 Rt = 1 x 4x 5 = 20 J

In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W, and 100 W are connected in parallel to and electrical source, then the bulb with the highest wattage glows with maximum brightness. Thus, the brightness of bulb B (100 W) is maximum. This brightness different bulks are in following order: B100 > B60 > B40

Resistivity depends on the nature of the material and the temperature but does not depend on the shape of the resistor.

Here, all these statements are correct. Hence, your answer will be option [D].

Power is given by, P = I²R P1 = I²R And, P2 = (2I)², R = 4I²R, When current is increased by 100% [It means that current becomes doubled i.e. 2I] Therefore, Increase in power dissipated = P2 – P1 = 4I²R – I²R Change in dissipated power = 3I^2R Increase in percent of power dissipated = 3P1/P1 ×100 = 300%

In first case, Resistivity of the conductor, ƿ = RA¹/ l In second case, Resistivity of the conductor, ƿ = RA¹/ 2l Therefore, for same material, resistivity of the conductor will be same. So, on substituting the value of ƿ in Eq. (ii), we have RA/I = RA¹/2I = A¹ =2A

Here, all these statements are correct. Hence, your answer will be option [D].

The minimum resistance is obtained when resistors are connected in parallel combination. This equivalent resistance will be: Rƿ = R/n = 1/5/5 = 1/25 Ω Rƿ = equivalent resistance for parallel combination.

The heating element of electric iron is made of nichrome wire which is a resistor. The electric wires are made of copper which is a good conductor but their covering is made of plastic which is an example of insulator.

Electrical resistivity of a given metallic wire depends on number density of free electrons in the conductor which is the nature of material. It also depends on the temperature of conductor.

If we take a lot of copper metal 1 metre long and 1 square metre in area of cross section then its resistance will be 1.69 x 10 to the power -8.

We know that, H = I²Rt It shows that the heat produced is proportional to the square of the current.

Here, all these statements are correct. Hence, your answer will be option [D].

This is so because in this case, the resistance of the wire will be low.

It is the temperature of an electric bulb filament when it glows.

The power in watts is found by multiplying the potential difference in volts by the current in amperes. We can write down the above formula for electric power in words as: Electric power = Potential difference x Current

Energy transferred, E = power x time = 100 W x 60 s = 6000 J

Electrical energy transferred per second, E = V x I x t = 6 V × 0.5 A × 1s = 3 J/s

This is so because in a parallel circuit, if one device fails, it does not affect the working of the other devices.

If the resistance wire is cut into five pieces, the resistance of each wire is R/5. If we connect the pieces in parallel, we will get the net resistance as R/25. Therefore, the ratio will be 25.

Resistance, R=ρlA Length, l = 300 m Cross-section area, A = 1.0 mm2 = 10-6 m2 Resistivity, ρ = 1.0 × 10−7 Ωm Resistance, R=10−7×30010−6 R = 30 Ω