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Physics MCQs for NEET — Practice Questions with Answers

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There are three resistance coils of equal resistance. The maximum number of resistances you can obtain by connecting them in any manner you choose, being free to use any number of the coils in any way is :

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Explanation

Maximum number of resistance $= 2^{n - 1} = 2^{3 - 1} = 4$

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Two wires of resistance R₁ and R₂ have temperature coefficient of resistance $α_{1} and α_{2}$ , respectively. These are joined in series. The effective temperature coefficient of resistance is :

$\frac{α_{1} + α_{2}}{2}$

$\sqrt{α_{1} α_{2}}$

$\frac{α_{1} R_{1} + α_{2} R_{2}}{R_{1} + R_{2}}$

$\frac{\sqrt{R_{1} R_{2} α_{1} α_{2}}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}$

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Explanation

$R_{t_{1}} = R_{1} (1 + α_{1} t)$ and $R_{t_{2}} = R_{2} (1 + α_{2} t)$

Also $R_{e q .} = R_{t_{1}} + R_{t_{2}} \Rightarrow R_{e q} = R_{1} + R_{2} + (R_{1} α_{1} + R_{2} α_{2}) t$

⇒ $R_{e q} = (R_{1} + R_{2}) \{1 + (\frac{R_{1} α_{1} + R_{2} α_{2}}{R_{1} + R_{2}}) . t\}$

So $α_{e f f} = \frac{R_{1} α_{1} + R_{2} α_{2}}{R_{1} + R_{2}}$

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When connected across the terminals of a cell, a voltmeter measures 5V and a connected ammeter measures 10 A of current. A resistance of 2 ohms is connected across the terminals of the cell. The current flowing through this resistance will be :

2.5 A

2.0 A

5.0 A

7.5 A

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Explanation

Emf E = 5V , Internal resistance $r = \frac{5}{10} = 0 .5 Ω$

Current through the resistance $i = \frac{5}{(2 + 0 .5)} = 2 A$

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Two resistances R₁ and R₂ are made of different materials. The temperature coefficient of the material of R₁ is α and of the material of R₂ is –β. The resistance of the series combination of R₁ and R₂ will not change with temperature, if R₁/ R₂ equals :

$\frac{α}{β}$

$\frac{α + β}{α - β}$

$\frac{α^{2} + β^{2}}{α β}$

$\frac{β}{α}$

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Explanation

$R_{1} + R_{2} = R_{1} (1 + α t) + R_{2} (1 - β t)$

⇒ $R_{1} + R_{2} = R_{1} + R_{2} + R_{1} α t - R_{2} β t$

⇒ $\frac{R_{1}}{R_{2}} = \frac{β}{α}$

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An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^{7}$ electrons and the same number of singly-charged positive ions per cm³. The electrons are moving at 0.4 m/s. The current density from anode to cathode is $4 μ A / m^{2}$ . The velocity of positive ions moving towards cathode is :

0.4 m/s

16 m/s

Zero

0.1 m/s

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Explanation

Current density of drifting electrons j = nev

$n = 5 \times 10^{7} c m^{- 3} = 5 \times 10^{7} \times 10^{6} m^{- 3}$ .

$v = 0 .4 m s^{- 1}, e = 1 .6 \times 10^{- 19} C$ ⇒ $j = 3 .2 \times 10^{- 6} A m^{- 2}$

Current density of ions = (4 – 3.2) × 10^–6 = $0 .8 \times 10^{- 6} \frac{A}{m^{2}}$

This gives v for ions = 0.1 ms^–1.

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A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to current, the temperature of the wire is raised by ΔT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross–section but of length 2 L. The temperature of the wire is raised by the same amount ΔT in the same time t. the value of N is

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Explanation

Let R and m be the resistance and mass of the first wire, then the second wire has resistance 2R and mass 2m. Let E = emf of each cell, S = specific heat capacity of the material of the wire. For the first wire, current $i_{1} = \frac{3 E}{R}$ and $i_{1}^{2} R t = m S Δ T$

For the second wire, $i_{2} = \frac{N E}{2 R}$ and $i_{2}^{2} (2 R) t = 2 m S Δ T$ . Thus, $i_{1} = i_{2}$ or $N = 6$ .

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The current in a conductor varies with time t as $I = 2 t + 3 t^{2}$ where I is in ampere and t in seconds. The electric charge flowing through a section of the conductor during t = 2 sec to t = 3 sec is :

10 C

24 C

33 C

44 C

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Explanation

dQ = Idt ⇒ $Q = \int_{t = 2}^{t = 3} I d t = [2 \int_{2}^{3} t d t + 3 \int_{2}^{3} t^{2} d t]$

= ${[t^{2}]}_{2}^{3} + {[t^{3}]}_{2}^{3}$ = (9 – 4) + (27 – 8) = 5 + 19 = 24C.

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Length of a hollow tube is 5m, it’s outer diameter is 10 cm and thickness of it’s wall is 5 mm. If the resistivity of the material of the tube is 1.7 × 10^–8 Ω×m then the resistance of the tube will be :

5.6 × 10^–5 Ω

2 × 10^–5 Ω

4 × 10^–5 Ω

None of these

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The resistance of the series combination of two resistance is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum possible value of n is :

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Explanation

If two resistances are $R_{1}$ and $R_{2}$ then

$S = R_{1} + R_{2}$ and $P = \frac{R_{1} R_{2}}{(R_{1} + R_{2})}$

From given condition S = nP i.e. $(R_{1} + R_{2}) = n (\frac{R_{1} R_{2}}{R_{1} + R_{2}})$

⇒ ${(R_{1} + R_{2})}^{2} = n R_{1} R_{2}$ ⇒ ${(R_{1} - R_{2})}^{2} + 4 R_{1} R_{2} = n R_{1} R_{2}$

So $n = 4 + \frac{{(R_{1} - R_{2})}^{2}}{R_{1} R_{2}} .$

Hence minimum value of n is 4.

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The V-I graph for a conductor makes an angle θ with V-axis. Here V denotes the voltage and I denotes current. The resistance of the conductor is given by :

$\sin θ$

$\cos θ$

$\tan θ$

$\cot θ$

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Explanation

At an instant approach the student will choose tanθ will be the right answer. But it is to be seen here the curve makes the angle θ with the V-axis. So it makes an angle (90 – θ) with the i-axis.

So resistance = slope = tan (90 – θ) = cotθ.

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