Class 10 Science Chapter 11 MCQ

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

In static electricity, the electric charges remain at rest or stationary, they do not move. The charge acquired by a glass rod rubbed with silk cloth and the charge acquired by an ebonite rod rubbed with woolen cloth are example of static electricity.

Power = 1 Kw, In Watt, P= 1000 W Voltage, V = 220 V Current, I = ? Current is given by I = P/V I= 1000/220 I = 4.5 A So, A fuse wire of rating slightly greater than 4.5A must be used i.e. 5A.

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

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

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

The current is directly proportional to the potential difference applied across the ends of a conductor it means that if the potential difference across the ends of a conductor is doubled, the current flowing through it also gets doubled, and if the potential difference is halved the current also gets halved.

As we know that: W= charge × potential difference W= Q × V As we know that Q= I × t Therefore, putting in above equation W= I × t × V V= Work done/ Current × time

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

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.

Given, Current, I = 1 A Time, t = 16 seconds No. of electrons, N = ? We know that, Current, I = Q/t = ne/t [e = charge of an electron = 1.6x 10 to the power -19 C and Q = ne] N= It/e = 1 x 16/ 1.6 x 10 to the power -19 = 10 to the power 20

We know that a thick wire has a greater area of cross section; whereas a thin wire has a smaller area of cross section. This means that a thick wire has less resistance and a thin wire has more resistance.

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.

An electric fuse works on the heating effect of current.

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

In series circuit, the appliances do not get the same voltage as that of the power supply line because the voltage is shared by all the appliances. The appliances get less voltage and hence do not work properly. For example, all the bulbs connected in series do not get the same voltage of 220 volts of the power supply line. They get less voltage and hence glow less brightly.

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

Kilowatt-hour is the commercial unit of electrical energy.

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

A power rating of 100 W on the bulb means that it will consume electrical energy at the rate of 100 joules per second. If we know the power P and voltage V of an electrical appliance, then we can very easily find out the current I drawn by it. This can be done by using the formula: P = V x I

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.

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