Class 10 Science Chapter 11 MCQ

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

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.

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].

Current is measured by an instrument called ammeter. The ammeter is always connected in series with the circuit in which the current is to be measured.

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].

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

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

The maximum resistance is obtained when resistors are connected in parallel combination. This equivalent resistance will be: Rs = n x R = 5 X 1/5 = 1Ω Rs = equivalent resistance for series combination.

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.

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.

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.

An electric fuse works on the heating effect of 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.

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

Current drawn by the kettle, I = PV=3000230 =13 A So, the cable should be able to carry more current than this and hence the answer is 15 A.

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