A cylindrical conductor of length
Question: A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross-section. (a)A/2 (b)3A/2 (c)2A (d)3A Solution: (c). Explanation :$\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$$\ldots(i)$ Also $\mathrm{R}=\frac{\rho(2 l)}{\mathrm{A}^{\prime}}$ $\ldots(i i)$ From eqns. $(i)$ and $(i i)$ $\frac{1}{\mathrm{~A}}=\frac{2}{\mathrm{~A}^{\prime}}$ or $\mathrm{A}^{\prime}=2 \mathrm{~A}$...
Read More →Ratna obtained a loan of Rs 25000 from the Syndicate Bank to renovate her house.
Question: Ratna obtained a loan of Rs 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt? Solution: Principal for the first year $=$ Rs. 25000 Interest for the first year $=$ Rs. $\left(\frac{25000 \times 8 \times 1}{100}\right)=$ Rs. 2000 Amount at the end of the first year $=$ Rs. $(25000+2000)=$ Rs. 27000 Principal for the second year $=$ Rs. 27000 Interest for the secon...
Read More →Find the difference between the simple interest and the compound interest on Rs 5000
Question: Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum. Solution: Principal amount $=$ Rs. 5000 Simple interest $=$ Rs. $\left(\frac{5000 \times 2 \times 9}{100}\right)=$ Rs. 900 The compound interest can be calculated as follows : Principal for the first year $=$ Rs. 5000 Interest for the first year $=$ Rs. $\left(\frac{5000 \times 9 \times 1}{100}\right)=$ Rs. 450 Amount at the end of the first year $=$ Rs. $(5000+450)=$ Rs. 5...
Read More →Using the principle of mathematical induction, prove each of the following
Question: Using the principle of mathematical induction, prove each of the following for all $\mathrm{n} \in \mathrm{N}$ : $2+6+18+\ldots+2 \times 3^{n-1}=\left(3^{n}-1\right)$ Solution: To Prove: $2+6+18+\ldots+2^{\times} 3^{n-1}=\left(3^{n}-1\right)$ Steps to prove by mathematical induction: Let $P(n)$ be a statement involving the natural number $n$ such that (i) $P(1)$ is true (ii) $P(k+1)$ is true, whenever $P(k)$ is true Then $P(n)$ is true for all $n \in N$ Therefore, Let $\mathrm{P}(\math...
Read More →Find the amount and the compound interest on Rs 15625 for 3 years at 12% per annum,
Question: Find the amount and the compound interest on Rs 15625 for 3 years at 12% per annum, compounded annually. Solution: Principal for the first year $=$ Rs. 15625 Interest for the first year $=$ Rs. $\left(\frac{15625 \times 12 \times 1}{100}\right)=$ Rs. 1875 Amount at the end of the first year $=$ Rs. $(15625+1875)=$ Rs. 17500 Principal for the second year $=$ Rs. 17500 Interest for the second year $=$ Rs. $\left(\frac{17500 \times 12 \times 1}{100}\right)=$ Rs. 2100 Amount at the end of ...
Read More →Which of the following represents voltage ?
Question: Which of the following represents voltage ? (a) $\frac{\text { Work done }}{\text { current } \times \text { time }}$ (b) Work done $\times$ charge (c) $\frac{\text { Work donex time }}{\text { current }}$ (d) Work done $\times$ charge $\times$ time Solution: (a). Explanation: Voltage or electric potential $=\frac{\text { Work done }}{\text { charge }}=\frac{\text { Work done }}{\text { current } \times \text { time }}$ ( $\because$ Current $=$ Charge $/$ time $)$...
Read More →Find the amount and the compound interest on Rs 2500 for 2 years at 10% per annum,
Question: Find the amount and the compound interest on Rs 2500 for 2 years at 10% per annum, compounded annually. Solution: Principal for the first year $=$ Rs. 2500 Interest for the first year $=$ Rs. $\left(\frac{2500 \times 10 \times 1}{100}\right)=$ Rs. 250 Amount at the end of the first year $=$ Rs. $(2500+250)=$ Rs. 2750 Principal for the second year $=$ Rs. 2750 Interest for the second year $=$ Rs. $\left(\frac{2750 \times 10 \times 1}{100}\right)=$ Rs. 275 Amount at the end of the second...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$ Solution: Let $y=\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$ $\Rightarrow y=a+b$ where $\mathrm{a}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{\mathrm{x}} ; \mathrm{b}=\mathrm{x}^{\left(1+\frac{1}{\mathrm{x}}\right)}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$ $\left\{\right.$ Using chain ...
Read More →The proper representation of series
Question: The proper representation of series combination of cells obtaining maximum potential is (a)(i) (b)(ii) (c)(iii) (d)(iv) (CBSE 2011,2012) Solution: (d). Explanation :Cells are connected in series if -ve terminal of one cell is connected to the +ve terminal of other cell....
Read More →Write 'T' for true and 'F' for false for each of the following:
Question: Write 'T' for true and 'F' for false for each of the following: (ii) $\mathrm{CP}=\frac{100}{(100+\text { gain } \%)} \times \mathrm{SP}$. (iii) Gain is reckoned on the selling price. (iv) The discount is allowed on the marked price. Solution: (i) False (F) $\mathrm{SP}=\left\{\frac{(100-\text { loss } \%)}{100} \times \mathrm{CP}\right\}$ (ii) True (T) (iii) False (F) Gain is reckoned on the cost price. (iv) True (T)...
Read More →Using the principle of mathematical induction, prove each of the following
Question: Using the principle of mathematical induction, prove each of the following for all n ϵ N: $1+3+3^{2}+3^{3}+\ldots+3^{n-1}=\frac{1}{2}\left(3^{n}-1\right)$ Solution: To Prove: $1+3^{1}+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$ Steps to prove by mathematical induction: Let $P(n)$ be a statement involving the natural number $n$ such that (i) $P(1)$ is true (ii) $P(k+1)$ is true, whenever $P(k)$ is true Then $P(n)$ is true for all $n \in N$ Therefore, Let $\mathrm{P}(\mathrm{n}): 1+3^{1}+3^{...
Read More →Fill in the blanks.
Question: Fill in the blanks. (i) The discount is reckoned on the ......... price. (ii) Gain or loss is always reckoned on the ......... (iii) SP = (Marked price) (.........) (iv) VAT is charged on the ......... of the article. Solution: (i) The discount is reckoned on themarkedprice. (ii) Gain or loss is always reckoned on thecost price. (iii) SP = (Marked price) (Discount). (iv) VAT is charged on theselling price of the article....
Read More →What is the minimum resistance
Question: What is the minimum resistance which can be made using 1 five resistors each of 1/5 Ω ? (a) $\frac{1}{5} \Omega$ (b) $25 \Omega$ (c) $\frac{1}{10} \Omega$ (d) $\frac{1}{25} \Omega$ Solution: (d). Explanation: $\frac{1}{R}=\frac{1}{1 / 5}+\frac{1}{1 / 5}+\frac{1}{1 / 5}+\frac{1}{1 / 5}+\frac{1}{1 / 5}=25$ or $R=\frac{1}{25} \Omega$...
Read More →Mark (✓) against the correct answer:
Question: Mark (✓) against the correct answer: The price of a watch including 8% VAT is Rs 810. What is its basic price? (a) Rs 675 (b) Rs 729 (c) Rs 750 (d) Rs 745 Solution: (c) Rs.750 Let the original price be Rs $x$. $\mathrm{VAT}=8 \%$ of Rs $x$ $=\operatorname{Rs}\left(x \times \frac{8}{100}\right)$ $=\operatorname{Rs} \frac{8 x}{100}$ $\therefore$ Price including VAT $=\mathrm{Rs}\left(x+\frac{8 x}{100}\right)$ $=R s . \frac{108 x}{100}$ Now, $\frac{108 x}{10}=810$ $\Rightarrow x=\left(81...
Read More →What is the maximum resistance
Question: What is the maximum resistance which can be made using five resistors each 1/5 Ω ? (a)10Ω (b)10 Ω (c)5Ω (d)I Ω (CBSE 2010) Solution: (d). Explanation: $R=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1 \Omega$...
Read More →Mark (✓) against the correct answer:
Question: Mark (✓) against the correct answer: A dealer lists his goods at 20% above cost price and allows a discount of 10%. His gain per cent is (a) 10% (b) 9%(c) 8% (d) 12% Solution: (c) 8% Let the CP be Rs 100 . Then, marked price $=$ Rs 120 Discount $=10 \%$ of $\mathrm{MP}$ $=(10 \%$ of Rs. 120$)$ $=$ Rs. $\left(120 \times \frac{10}{100}\right)$ $=$ Rs. 12 Now, SP $=($ MP $)-($ discount $)$ $=$ Rs $(120-12)$ $=$ Rs 108 $\therefore$ Gain percentage $=(108-100) \%$ $=8 \%$...
Read More →Using the principle of mathematical induction, prove each of the following
Question: Using the principle of mathematical induction, prove each of the following for all $n \in \bar{N}$ : 2 + 4 + 6 + 8 + . + 2n = n(n + 1) Solution: To Prove: 2 + 4 + 6 + 8 + . + 2n = n(n + 1) Steps to prove by mathematical induction: Let $P(n)$ be a statement involving the natural number $n$ such that (i) $P(1)$ is true (ii) $P(k+1)$ is true, whenever $P(k)$ is true Then $P(n)$ is true for all $n \in N$ Therefore, Let $P(n): 2+4+6+8+\ldots .+2 n=n(n+1)$ Step 1: P(1) = 1(1 + 1) = 1 2 = 2 T...
Read More →Identify the circuit (Figure) in which electrical
Question: Identify the circuit (Figure) in which electrical components have been properly connected. (a)(i) (b)(ii) (c)(iii) (d)(iv) (CBSE 2010) Solution: (b). Explanation :Voltmeter is always connected in parallel and ammeter is connected in series in an electric circuit. Moreover, positive terminals of voltmeter and ammeter are connected with the +ve terminal of a cell or battery and negative terminals of voltmeter and ammeter are connected with -ve terminal of a cell or battery....
Read More →A current of 1A is drawn by a filament of an electric bulb.
Question: A current of 1A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly (a)1020 (b)1016 (c)1018 (d)1023 Solution: (a). Explanation : $I=\frac{Q}{t}=\frac{n e}{t}$ $\therefore$ $n=\frac{\mathrm{It}}{e}=\frac{1 \times 16}{1.6 \times 10^{-19}}=10^{20}$...
Read More →Mark (✓) against the correct answer:
Question: Mark (✓) against the correct answer: On selling a chair for Rs 680, a man loses 15%. To gain 15%, it must be sold for (a) Rs 800 (b) Rs 860 (c) Rs 920 (d) Rs 884 Solution: (c) Rs 920 $\mathrm{SP}=\mathrm{Rs} 680$ Loss percentage $=15 \%$ Now, $\mathrm{CP}=\left\{\frac{100}{(100-\text { loss } \%)} \times \mathrm{SP}\right\}$'$=$ Rs. $\left\{\frac{100}{(100-15)} \times 680\right\}$ $=$ Rs. $\left(\frac{100}{85} \times 680\right)$ $=$ Rs. 800 $\therefore$ Desired SP $=\left\{\frac{(100+\...
Read More →Electrical resistivity of a given metallic
Question: Electrical resistivity of a given metallic wire depends upon (a)its length (b)its thickness (c)its shape (d)nature of material. Solution: (d). Explanation :Resistivity of a given metallic wire does not depend on its dimensions and shape but depends on the nature of the material of the wire....
Read More →Mark (✓) against the correct answer
Question: Mark (✓) against the correct answer The selling price of an article is $\frac{6}{5}$ of the cost price. The gain per cent is (a) 15% (b) 20% (c) 25% (d) 30% Solution: (b) 20% Let $x$ be the CP of the article. Then, SP of the article $=\frac{6}{5} x$ Now, gain $=(\mathrm{SP}-\mathrm{CP})$ $=\left(\frac{6}{5} \mathrm{x}-\mathrm{x}\right)$ $=\frac{\mathrm{x}}{5}$ $\therefore$ Gain percentage $=\left(\frac{\text { gain }}{\mathrm{CP}} \times 100\right) \%$ $=\left(\frac{\frac{x}{5}}{x} \ti...
Read More →In the following circuits shown in figure,
Question: In the following circuits shown in figure, power produced in the resistor or combination of resistors connected to a 12V battery will be (a)same in all cases (b)minimum in case (i) (c)maximum in case (ii) (d)maximum in case (iii) Solution: (d). Explanation: Power produced in case $(i)$ $=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{144}{2}=77 \mathrm{~J} \mathrm{~s}^{-1}$ Power produced in case $(i i)=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{144}{4}=36 \mathrm{Js}^{-1}$ Power produced in cas...
Read More →Using the principle of mathematical induction, prove each of the following
Question: Using the principle of mathematical induction, prove each of the following for all n ϵ N: $1+2+3+4+\ldots+n=1 / 2 n(n+1)$ Solution: To Prove: $1+2+3+4+\ldots+n=1 / 2 n(n+1)$ Steps to prove by mathematical induction: Let $P(n)$ be a statement involving the natural number $n$ such that (i) $P(1)$ is true (ii) $P(k+1)$ is true, whenever $P(k)$ is true Then $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$ Therefore, Let $P(n): 1+2+3+4+\ldots+n=1 / 2 n(n+1)$ Step 1: P(1)...
Read More →A cell, a resistor, a key and ammeter
Question: A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of the figure given below. The current recorded in the ammeter will be (a)maximum in (i) (b)maximum in (ii) (c)maximum in (iii) (d)the same in all the cases. Solution: (d).Explanation :Same current flows through every part of the circuit having resistance connected in series to a cell or battery....
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