At what rate per cent per annum will Rs 640 amount to Rs 774.40 in 2 years
Question: At what rate per cent per annum will Rs 640 amount to Rs 774.40 in 2 years when compounded annually? Solution: Let the required rate be $R \% \mathrm{p}$. a. $A=774.40$ $P=640$ $\mathrm{n}=2$ years Now, $A=P\left(1+\frac{R}{100}\right)^{n}$ $\Rightarrow 774.40=640\left(1+\frac{R}{100}\right)^{2}$ $\Rightarrow \frac{774.40}{640}=\left(1+\frac{R}{100}\right)^{2}$ $\Rightarrow 1.21=\left(1+\frac{R}{100}\right)^{2}$ $\Rightarrow(1.1)^{2}=\left(1+\frac{R}{100}\right)^{2}$ $\Rightarrow 1.1-1...
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: $3 \cdot 2^{2}+3^{2} \cdot 2^{3}+3^{3} \cdot 2^{4}+\ldots .+3^{n} \cdot 2^{n+1}=\frac{12}{5}\left(6^{n}-1\right)$ Solution: To Prove: $3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{n} \times 2^{n+1}=\frac{12}{5}\left(6^{n}-1\right)$ Let us prove this question by principle of mathematical induction (PMI) Let $\mathrm{P}(\mathrm{n}): 3 \times 2^{2}+3^{2} \times 2^{3}+3...
Read More →Find the values
Question: Find $\frac{d y}{d x}$, when $y=x^{n}+n^{x}+x^{x}+n^{n}$ Solution: let $y=x^{n}+n^{x}+x^{x}+n^{n}$ $\Rightarrow y=a+b+c+m$ where $a=x^{n} ; b=n^{x} ; c=x^{x} ; m=n^{n}$ $\frac{d y}{d x}=\frac{d a}{d x}+\frac{d b}{d x}+\frac{d c}{d x}+\frac{d m}{d x}$ $\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where $\mathrm{a}$ and $\mathrm{u}$ are any variables $\}$ $a=x^{n}$ Taking log both...
Read More →Three incandescent bulbs of 100W
Question: Three incandescent bulbs of 100W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same voltage are connected in parallel to the same source. (a) Will the bulb in the two circuits glow with the same brightness ? Justify your answer. (b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit ? Give reason. (CBSE 2012) Solution: (a) Power dissipated in a circuit = V2/R. Since re...
Read More →At what rate per cent per annum will Rs 4000 amount to Rs 4410 in 2 years
Question: At what rate per cent per annum will Rs 4000 amount to Rs 4410 in 2 years when compounded annually? Solution: Let $R \% p . a .$ be the required rate. $A=4410$ $P=4000$ $n=2$ years Now,$A=\mathrm{P}\left(1+\frac{R}{100}\right)^{n}$ $\Rightarrow 4410=4000\left(1+\frac{\mathrm{R}}{100}\right)^{2}$ $\Rightarrow \frac{4410}{4000}=\left(1+\frac{R}{100}\right)^{2}$ $\Rightarrow \frac{441}{400}=\left(1+\frac{R}{100}\right)^{2}$ $\Rightarrow\left(\frac{21}{20}\right)^{2}=\left(1+\frac{R}{100}\...
Read More →B1, B2 and B3 are three identical bulbs connected as shown in figure.
Question: B1, B2and B3are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A. What happenes to the glow of other bulbs when the bulb B1gets fused ? What happens to the reading of A1, A2, A3and A when the bulb B2gets fused ? How much power is dissipated in the circuit when all the three bulbs glow together ? Solution: The glow of bulb depends upon the energy disspated per second i.e. P =V2/R.Since V and R of both the bul...
Read More →What sum of money will amount to Rs 21296 in 3 years at 10%
Question: What sum of money will amount to Rs 21296 in 3 years at 10% per annum, compounded annually? Solution: Let $P$ be the sum. Rate of interest, $R=10 \%$ Time, $n=3$ years Now, $A=P \times\left(1+\frac{10}{100}\right)^{3}$ $=$ Rs. $P \times\left(\frac{100+10}{100}\right)^{3}$ $=$ Rs. $P \times\left(\frac{110}{100}\right)^{3}$ $=$ Rs. $P \times\left(\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}\right)$ $=$ Rs. $\frac{1331 P}{1000}$ However, amount $=$ Rs. 21296 Now, Rs. $21296=$ R...
Read More →A sum of money amounts to Rs 10240 in 2 years
Question: A sum of money amounts to Rs 10240 in 2 years at $6 \frac{2}{3} \%$ per annum, compounded annually. Find the sum. Solution: Let $P$ be the sum. Rate of interest, $R=6 \frac{2}{3} \%=\frac{20}{3} \%$ Time, $n=2$ years Now, $A=P \times\left(1+\frac{20}{100 \times 3}\right)^{2}$ $=$ Rs. $P \times\left(1+\frac{20}{300}\right)^{2}$ $=$ Rs. $P \times\left(\frac{300+20}{300}\right)^{2}$ $=$ Rs. $P \times\left(\frac{320}{300}\right)^{2}$ $=$ Rs. $P \times\left(\frac{16}{15} \times \frac{16}{15...
Read More →The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs 93.
Question: The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs 93. Find the sum. Solution: Let $P$ be the sum. Then SI $=$ Rs $\left(\frac{P \times 3 \times 10}{100}\right)=$ Rs $\frac{30 P}{100}=$ Rs $\frac{3 P}{10}$ Also, CI $=$ Rs. $\left\{P \times\left(1+\frac{10}{100}\right)^{3}-P\right\}$ $=$ Rs. $\left\{P \times\left(\frac{100+10}{100}\right)^{3}-P\right\}$ $=$ Rs. $\left\{P \times\left(\frac{11}{10}\right)^{3}-P\right\}$...
Read More →Why is parallel arrangement used
Question: Why is parallel arrangement used in domestic wiring ? (CBSE 2012) Solution: If any one of the electric devices in parallel fuses, then the working of other devices will not be affected. When different devices are connected in parallel, they draw the current as per their requirement and hence they work properly....
Read More →A current of 1 ampere flows in a series circuit containing
Question: A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω when connected to a 10V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω condcutor and potentional difference across the lamp will take place ? Give reason. Draw circuit diagram. (CBSE 2010) Solution: Let $\quad R_{1}=$ resistance of electric lamp. Tot...
Read More →The difference between the compound interest and the simple interest on a certain sum for 2 years
Question: The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs 90. Find the sum. Solution: Let $\operatorname{Rs} P$ be the sum. Then SI $=\left(\frac{P \times 2 \times 6}{100}\right)=$ Rs. $\frac{12 P}{100}=$ Rs. $\frac{3 P}{25}$ Also, CI $=\left\{P \times\left(1+\frac{6}{100}\right)^{2}-P\right\}$ $=$ Rs. $\left\{P \times\left(\frac{100+6}{100}\right)^{2}-P\right\}$ $=$ Rs. $\left\{P \times\left(\frac{53}{50}\right)^{2}-P\right...
Read More →The simple interest on a sum of money for 2 years at 8% per annum is Rs 2400.
Question: The simple interest on a sum of money for 2 years at 8% per annum is Rs 2400. What will be the compound interest on that sum at the same rate and for the same period? Solution: Simple interest $(\mathrm{SI})=$ Rs. 2400 Rate of interest, $R=8 \%$ Time, $n=2$ years The principal can be calculated using the formula: $\operatorname{Sum}=\left(\frac{100 \times \mathrm{SI}}{\mathrm{R} \times \mathrm{T}}\right)$ $\Rightarrow$ Sum $=$ Rs. $\left(\frac{100 \times 2400}{8 \times 2}\right)=$ Rs. ...
Read More →What is the commercial unit
Question: What is the commercial unit of electrical energy ? Represent it in terms of joule. Solution: kilowatt hour (kWh) is the commercial unit of electrical energy. $1 \mathrm{kWh}=1000 \mathrm{Watt} \times 3600 \mathrm{~s}$ $=3.6 \times 10^{6} \frac{\mathrm{J}}{\mathrm{s}} \times \mathrm{s}$ $=3.6 \times 10^{6} \mathrm{~J}(\because 1 \mathrm{Watt}=1 \mathrm{~J} / \mathrm{s})$...
Read More →Find the values
Question: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when $1 y=e^{x}+10^{x}+x^{x}$ Solution: let $y=e^{x}+10^{x}+x^{x}$ $\Rightarrow y=a+b+c$ where $\mathrm{a}=\mathrm{e}^{\mathrm{x}} ; \mathrm{b}=10^{\mathrm{x}} ; \mathrm{c}=\mathrm{x}^{\mathrm{x}}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}+\frac{\mathrm{dc}}{\mathrm{dx}}$ $\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+...
Read More →In a series electrical circuit comprising
Question: In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5A. The reading of ammeter decreases to half when the length of the wire is doubled. Why ? Solution: Resistance of wire length $l, \mathrm{R}=\frac{\rho l}{\mathrm{~A}}$. When length becomes double, $\mathrm{R}^{\prime}=\frac{2 \rho l}{\mathrm{~A}}$. $\therefore \quad \frac{\mathrm{R}^{\prime}}{\mathrm{R}}=2 \quad$ or $\quad \mathrm{R}^{\prime}=2 \mathrm{R}$ $I=\frac{V}{R} \quad$ and $I^{...
Read More →Abhay borrowed Rs 16000 at
Question: Abhay borrowed Rs 16000 at $7 \frac{1}{2} \%$ per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years? Solution: Principal amount, $P=$ Rs 16000 Rate of interest, $R=\frac{15}{2} \%$ p. a. Time, $n=2$ years Now, simple interest $=$ Rs $\left(\frac{16000 \times 2 \times 15}{100 \times 2}\right)=$ Rs. 2400 Amount including the simple interest $=$ Rs $(16000+2400)=$ Rs 18400 The formula for $t h e$...
Read More →How does use of a fuse wire protect
Question: How does use of a fuse wire protect electrical appliances ?(CBSE 2012) Solution: When large current flows in an electric circuit, fuse wire melts due to the large heat produced in it. Therefore, current stops flowing in the circuit and electrical appliances connected in the circuit are protected from burning due to large current in j the circuit....
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: $1.2+2.2^{2}+3.2^{3}+\ldots . .+n .2^{n}=(n-1) 2^{n+1}+2$ Solution: To Prove: $1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+\ldots \ldots+n \times 2^{n}=(n-1) 2 n+1+2$ Let us prove this question by principle of mathematical induction (PMI) Let $\mathrm{P}(\mathrm{n}): 1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+\ldots \ldots+n \times 2^{n}$ For $n=1$ LHS $=1 \times 2=2$ RHS $=(1-1) \times 2^{(1+1...
Read More →Draw a circuit diagram of an electric circuit containing a cell,
Question: Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a reistor of 2Ω in series with a combination of two resistors (4Ω each) in parallel and a voltmeter across the energy. 1 kWh = 1000 Watt x 3600 s parallel combination. Will the potentional difference across the 2Ω resistor be the same as that across the parallel combination of 4Ω resistors ? Give reason. Solution: $\frac{1}{R}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ or $R=2 \Omega$ Therefore, the above cir...
Read More →Neha borrowed Rs 24000 from the State Bank of India to buy a scooter.
Question: Neha borrowed Rs 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months? Solution: Principal amount, $P=$ Rs. 24000 Rate of interest, $R=10 \%$ p. a. Time, $n=2$ years 3 months $=2 \frac{1}{4}$ years The formula for the amount including the compound interest is given below: $\mathrm{A}=P \times\left(1+\frac{R}{100}\right)^{n} \times\left(1+\frac{\frac{1}{4} R}{100}\righ...
Read More →Shubhalaxmi took a loan of Rs 18000 from surya Finance to purchase a TV set.
Question: Shubhalaxmi took a loan of Rs 18000 from surya Finance to purchase a TV set. If the company charges compound interest at $12 \%$ per annum during the first year and $12 \frac{1}{2} \%$ per annum during the second year, how much will she have to pay after 2 years? Solution: Principal amount, $P=$ Rs. 18000 Rate of interest for the first year, $p=12 \%$ p. a. Rate of interest for the second year, $q=12 \frac{1}{2} \%$ p.a. Time, $n=2$ years The formula for the amount including the compou...
Read More →Should the resistance of an ammeter
Question: Should the resistance of an ammeter be low or high ? Give reason. Solution: Ammeter is connected in series in an electric circuit to measure electric current. If its resistance is high, then the net resistance of the electric circuit will increase and hence current in the electric circuit will decrease. Hence, ammeter will not read the actual value of the current in the circuit. If resistance of the ammeter is low, then the net resistance of the circuit will not be effected much. Hence...
Read More →Three 20 resistors, A, B and C are connected as shown in figure.
Question: Three 20 resistors, A, B and C are connected as shown in figure. Each of them dissipates energy that can withstand a maximum power of 18W without melting. Find the maximum current that can flow through three resistors. (CBSE, 2010) Solution: Here, $P=18 W$ Using, $P=I^{2} R$, we get $I=\sqrt{\frac{\mathrm{P}}{\mathrm{R}}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3 \mathrm{~A}$ $\therefore$ Maximum current that can flow through a resistor $A=3 A$ Since resistors B and C are connected in parallel, s...
Read More →Three years ago, Beeru purchased a buffalo from Surjeet for Rs 11000.
Question: Three years ago, Beeru purchased a buffalo from Surjeet for Rs 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually? Solution: Principal amount, $P=$ Rs. 11000 Rate of interest, $R=10 \%$ p. a. Time, $n=3$ years The amount including the compound interest is calculated using the formula, $A=$ Rs. $P\left(1+\frac{R}{100}\right)^{n}$ $=$ Rs. $11000\left(1+\frac{10}{100}\right)^{3}$ $=$ Rs. $11000\left(\frac{100+10}{100}\right)^{3}$...
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