CBSE 10th Standard Science Subject Electricity Chapter Case Study Questions With Solution 2021
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CBSE 10th Standard Science Subject Electricity Case Study Questions With Solution 2021
10th Standard CBSE
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Reg.No. :
Science
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The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current.
The electric potential is defined as the amount of work done in bringing a unit positive test charge from infinity to a point in the electric field. The amount of work done in bringing a unit positive test charge from one point to another point in an electric field is defined as potential difference.
\(\begin{equation} V_{A B}=V_{B}-V_{A}=\frac{W_{B A}}{q} \end{equation}\)
The SI unit of potential and potential difference is volt.
(i) The 2 C of charge is flowing through a conductor in 100 rns, the current in the circuit is(a) 20 A (b) 2 A (c) 0.2 A (d) 0.02 A (ii) Which of the following is true?
(a) Current flows from positive terminal of the cell to the negative terminal of the cell outside the cell.
(b) The negative charge moves from lower potential to higher potential.
(c) The direction of flow of current in same as the direction of flow of positive charge.
(d) All of these
(iii) The potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to other is(a) 50 V (b) -5 V (c) 0.5 V (d) 500 V (iv) The number of electrons flowing per second in a conductor if 1A current is passing through it
(a) 6.25 x 1020 (b) 6.25 x 1019 (c) 6.25 X 1018 (d) 6.25 x 10-19 (v) The voltage can be written as
(a) Work done x charge x time (b) \(\begin{equation} \frac{\text { Work done }}{\text { Current } \times \text { time }} \end{equation}\) (c) \(\begin{equation} \frac{\text { Work done } \times \text { time }}{\text { Current }} \end{equation}\) (d) Work done x charge -
The relationship between potential difference and current was first established by George Simon Ohm called Ohm's law. According to this law, the current through a metallic conductor is proportional to the potential difference applied between its ends, provided the temperature remain constant i.e. I \(\begin{equation} \propto \end{equation}\) V or V = IR; where R is constant for the conductor and it is called resistance of the conductor. Although Ohm's law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold.
(i) If both the potential difference and the resistance in a circuit are doubled, then(a) current remains same (b) current is doubled (c) current is halved (d) current is quadrupled (ii) For a conductor, the graph between V and I is there. Which one is the correct?
(a) 
(b) 
(c) 
(d) 
(iii) The slope of V - I graph (V on x-axis and I on y-axis) gives
(a) resistance (b) reciprocal of resistance (c) charge (d) reciprocal of charge. (iv) When battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is
(a) 9 \(\begin{equation} \Omega \end{equation}\) (b) 0.9 \(\begin{equation} \Omega \end{equation}\) (c) 90 \(\begin{equation} \Omega \end{equation}\) (d) 900 \(\begin{equation} \Omega \end{equation}\) (v) By increasing the voltage across a conductor, the
(a) current will decrease (b) resistance will increase (c) current will increase (d) resistance will decrease. -
The obstruction offered by a conductor in the path of flow of current is called resistance. The SI unit of resistance is ohm (\(\begin{equation} \Omega \end{equation}\)). It has been found that the resistance of a conductor depends on the temperature of the conductor. As the temperature increases the resistance also increases. But the resistance of alloys like mangnin, constantan and nichrome is almost unaffected by temperature. The resistance of a conductor also depends on the length of conductor and the area of cross-section of the conductor. More be the length, more will be the resistance, more be the area of cross-section, lesser will be the resistance.
(i) Which of the following is not will desired in material being used for making electrical wires?(a) High melting point (b) High resistance (c) High conductivity (d) None of these (ii) The V - I graph for two metallic wires A and B is given. What is the correct relationship between their temperatures?
(a) TA < TB (b) TA > TB (c) TA = TB (d) none of these (iii) Two wires of same material one of length L and area of cross-section A, other is of length 2L and area A/2 . Which of the following is correct?
(a) R1 = R2 (b) R1 = 4R2 (c) R2 = 4R1 (d) R1 = 2R2 (iv) For the same conducting wire
(a) resistance is higher in summer
(b) resistance is higher in winter
(c) resistance is same is summer or in winter
(d) none of these
(v) A wire of resistance 20 \(\begin{equation} \Omega \end{equation}\) is cut into 5 equal pieces. The resistance of each part is(a) 4 \(\begin{equation} \Omega \end{equation}\) (b) 10 \(\begin{equation} \Omega \end{equation}\) (c) 100 \(\begin{equation} \Omega \end{equation}\) (d) 80 \(\begin{equation} \Omega \end{equation}\) -
Two or more resistances are connected in series or in parallel or both, depending upon whether we want to increase or decrease the circuit resistance.
The two or more resistances are said to be connected in series if the current flowing through each resistor is same. The equivalent resistance in the series combination is given by
Rs = R1 + R2 + R3
(i) When three resistors are connected in series with a battery of voltage V and voltage drop across resistors is V1, V2 and V3, which of the relation is correct?(a) V = V1= V2 = V3 (b) V = V1 + V2 + V3 (c) V1 + V2 + V3 = 3V (d) V > V1 + V2 + V3 (ii) When the three resistors each of resistance R ohm, connected in series, the equivalent resistance is
(a) R/2 (b) > R (c) < R/2 (d) < R (iii) There is a wire oflength 20 cm and having resistance 20 \(\begin{equation} \Omega \end{equation}\) cut into 4 equal pieces and then joined in series.
The equivalent resistance is(a) 20 \(\begin{equation} \Omega \end{equation}\) (b) 4 \(\begin{equation} \Omega \end{equation}\) (c) 5 \(\begin{equation} \Omega \end{equation}\) (d) 10 \(\begin{equation} \Omega \end{equation}\) (iv) In the following circuit, find the equivalent resistance between A and B is (R = 2 \(\begin{equation} \Omega \end{equation}\))
(a) 10 \(\begin{equation} \Omega \end{equation}\) (b) 5 \(\begin{equation} \Omega \end{equation}\) (c) 2 \(\begin{equation} \Omega \end{equation}\) (d) 4 \(\begin{equation} \Omega \end{equation}\) (v) In the given circuit, the current in each resistor is
(a) 3 A (b) 6 A (c) 9 A (d) 18 A
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CBSE 10th Standard Science Subject Electricity Case Study Questions With Solution 2021 Answer Keys
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(i) (a): q = 2 C, t= 100 ms = 0.1 s
\(\begin{equation} I=\frac{q}{t}=\frac{2}{0.1}=20 \mathrm{~A} \text { . } \end{equation}\)
(ii) (d)
(iii) (b): W = 100 J, q = 20 C
\(\begin{equation} V=\frac{W}{q}=\frac{100}{20}=5 \mathrm{~V} \end{equation}\)
(iv) (c): I = 1 A, t = 1 s
q = It = 1 x 1 = 1C
\(\begin{equation} n=\frac{q}{e}=\frac{1}{1.6 \times 10^{-19}}=6.25 \times 10^{18} \end{equation}\)
(v) (c): \(\begin{equation} V=\frac{W}{q}=\frac{W}{I t} \end{equation}\) -
(i)(a): V = IR
So, V' \(\rightarrow\) 2 V, R' \(\rightarrow\) 2R
\(\begin{equation} I^{\prime}=\frac{2 V}{2 R}=I \end{equation}\)
(ii) (b): V \(\begin{equation} \propto \end{equation}\) I. So, the graph is a straight line and passing through origin.
(iii) (b): Slope of V-I graph = \(\begin{equation} \frac{I}{V}=\frac{1}{R} \text { . } \end{equation}\)
(iv) (c): Given: V = 9 V, 1= 0.1 A
\(\begin{equation} R=\frac{V}{I}=\frac{9}{0.1}=90 \Omega \end{equation}\)
(v) (c): On increasing the voltage, the resistance remain same, so current will increase . -
(i) (b): The electrical wire should have low resistance.
(ii) (b): More is the temperature, more will be the resistance. The resistance of A is more, so temperature of A is more.
(iii) (c): \(\begin{equation} R_{1}=\rho \frac{L}{A}, R_{2}=\rho \frac{2 L}{A / 2} \end{equation}\)
(iv) (a): In summers, temperature is more, so resistance is more.
(v) (a): \(\begin{equation} R^{\prime}=\frac{R}{5}=\frac{20}{5}=4 \Omega \end{equation}\) -
(i) (b): In series combination, the total voltage is equal to the sum of voltage drop across each resistance.
(ii) (b): Rs = R1 + R2 + R3
So, Rs = R + R + R = 3R
(iii) (a): Resistance of each wire = 20/4 = 5 \(\begin{equation} \Omega \end{equation}\)
Equivalent resistance in series
Rs = 5 + 5 + 5 + 5 = 20 \(\begin{equation} \Omega \end{equation}\)
(iv) (a): All are in series, Rs = 5R= 5 x 2 = 10\(\begin{equation} \Omega \end{equation}\)
(v) (a): Rs = 1 + 2 + 3 = 6 \(\begin{equation} \Omega \end{equation}\)
\(\begin{equation} I=\frac{18}{6}=3 \mathrm{~A} \end{equation}\)
