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Question: Find $\frac{d y}{d x}$, when $y=\sin x \sin 2 x \sin 3 x \sin 4 x$ Solution: Let $y=\sin x \sin 2 x \sin 3 x \sin 4 x$ Take log both sides: $\Rightarrow \log y=\log (\sin x \sin 2 x \sin 3 x \sin 4 x)$ $\Rightarrow \log y=\log (\sin x)+\log (\sin 2 x)+\log (\sin 3 x)+\log (\sin 4 x)$ $\left\{\log (a b)=\log a+\log b ; \log \left(\frac{a}{b}\right)=\log a-\log b\right\}$ Differentiating with respect to $x$ : $\Rightarrow \frac{d(\log y)}{d x}=\frac{d(\log (\sin x))}{d x}+\frac{d(\log (\...
Read More →Find the amount and the compound interest on Rs 8000 for 1 year
Question: Find the amount and the compound interest on Rs 8000 for 1 year at 10% per annum, compounded half-yearly. Solution: Principal, $P=$ Rs. 8000 Time, $n=1$ year $=2$ half years Rate of interest per annum $=10 \%$ Rate of interest for half year, $R=\frac{10 \%}{2}=5 \%$ The amount with the compound interest is given by Amount $=$ Rs. $P \times\left(1+\frac{R}{100}\right)^{2}$ $=$ Rs. $8000 \times\left(1+\frac{5}{100}\right)^{2}$ $=$ Rs. $8000 \times\left(\frac{105}{100}\right)^{2}$ $=$ Rs....
Read More →Choose the incorrect statement from
Question: Choose the incorrect statement from the following regarding magnetic lines of field. (a)The direction of magnetic field at a point is taken to be the direction in which the north pole of a magnetic compass needle points. (b)Magnetic field lines are closed curves. (c)If magnetic field lines are parallel and equidistant, they rep-resent zero field strength. (d)Relative strength of magnetic field is shown by the degree of closeness of the field lines. Solution: (c). Explanation :Magnetic ...
Read More →The value of a machine depreciates at the rate of 10% per annum.
Question: The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased? Solution: Let the initial value of the machine, $P$ be Rs $x$. Rate of depreciation, $R=10 \%$ Time, $n=3$ years The present value of the machine is Rs 291600 . Then the initial value of the machine is given by Value $=P \times\left(1-\frac{R}{100}\right)^{n}$ $=$ Rs. $x \times\left(1-\frac{10}{100}\right)^{3}$ $=$ Rs. $x \time...
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: $\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 n-1) \times(3 n+2)}=\frac{n}{(6 n+4)}$ Solution: To Prove: $\frac{1}{2 \times 5}+\frac{1}{(5 \times 8)}+\ldots \ldots+\frac{1}{(3 n-1) \times(3 n+2)}=\frac{n}{(6 n+4)}$ For n = 1 $\mathrm{LHS}=\frac{1}{2 \times 5}=\frac{1}{10}$ $\mathrm{RHS}=\frac{1 \times 1}{(6+4)}=\frac{1}{10}$ Hence, LHS = RHS $P(n)$ is true for $n=1$ Assume...
Read More →Solve this
Question: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when $y=e^{3 x} \sin 4 x 2^{x}$ Solution: Let $y=e^{3 x} \sin 4 x 2^{x}$ Take log both sides: $\Rightarrow \log y=\log \left(e^{3 x} \sin 4 x 2^{x}\right)$ $\Rightarrow \log y=\log \left(e^{3 x}\right)+\log (\sin 4 x)+\log \left(2^{x}\right)$ $\left\{\log (a b)=\log a+\log b ; \log \left(\frac{a}{b}\right)=\log a-\log b\right\}$ $\Rightarrow \log y=3 x \log e+\log (\sin 4 x)+x \log 2\left\{\log x^{a}=a \log x\right\}$ $\Rightarrow \log y=3 x+\log...
Read More →Find out the following in the circuit given in the figure.
Question: Find out the following in the circuit given in the figure. (a) Effective resistance of two 8Ω resistors in the combination. (b) Current flowing through 40 resistor (c) Potentional difference across 40 resistor (d) Power dissipated in 4Ω resistor (e) Difference in ammeter readings, if any. (CBSE 2010,2012) Solution: (a)Two 80 resistors are in parallel, so their effective resistance is given by $\frac{1}{\mathrm{R}}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ or $\mathrm{R}=4 \Omega$ (b) Net re...
Read More →A car is purchased for Rs 348000.
Question: A car is purchased for Rs 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years? Solution: Initial value of the car, $P=$ Rs 348000 Rate of depreciation for the first year, $p=10 \%$ Rate of depreciation for the second year, $q=20 \%$ Time, $n=2$ years. Then the value of the car after two years is given by Value $=\left\{P \times\left(1-\frac{p}{100}\right) \times\left(1-\frac{q}{100}\right...
Read More →A scooter is bought at Rs 56000.
Question: A scooter is bought at Rs 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years? Solution: Initial value of the scooter, $P=$ Rs 56000 Rate of depreciation, $R=10 \%$ Time, $n=3$ years Then the value of the scooter after three years is given by Value $=P \times\left(1-\frac{R}{100}\right)^{n}$ $=$ Rs. $56000 \times\left(1-\frac{10}{100}\right)^{3}$ $=$ Rs. $56000 \times\left(\frac{100-10}{100}\right)^{3}$ $=$ Rs. $56000 \times\left(\frac{90}{10...
Read More →A machine is purchased for Rs 625000.
Question: A machine is purchased for Rs 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years? Solution: Initial value of the machine, $P=$ Rs 625000 Rate of depreciation, $R=8 \%$ Time, $n=2$ years Then the value of the machine after two years is given by Value $=P \times\left(1-\frac{R}{100}\right)^{n}$ $=$ Rs $625000 \times\left(1-\frac{8}{100}\right)^{2}$ $=$ Rs $625000 \times\left(\frac{100-8}{100}\right)^{2}$ $=$ Rs $625000 \times\left(\frac{92}{10...
Read More →Solve this
Question: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when $y=\frac{e^{\operatorname{ex}} \operatorname{sex} x \log x}{\sqrt{1-2 x}}$ Solution: Let $y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{\sqrt{1-2 x}}$ $\Rightarrow y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{(1-2 x)^{\frac{1}{2}}}$ Take log both sides: $\Rightarrow \log y=\log \left(\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{(1-2 x)^{\frac{1}{2}}}\right)$ $\Rightarrow \log y=\log e^{2 x}+\log \sec ^{x} x+\log \log x-\log (1-2 ...
Read More →The bacteria in a culture grows by 10% in the first hour,
Question: The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000. Solution: Initial count of bacteria, $P=20000$ Rate of increase, $R=10 \%$ Time, $n=3$ hours Then the count of bacteria at the end of the first hour is given by Count of bacteria $=P \times\left(1+\frac{10}{100}\right)^{n}$ $=20000 \times\left(1+\frac{10}{100}\right)^{1}$ $=2...
Read More →What is Joule’s heating effect ?
Question: What is Joules heating effect ? How can it be demonstrated experimentally ? List its four applications in daily life. (CBSE 2012) Solution: Joules law can be stated as :The amount of heat produced in a conductor is (i)directly proportional to the square of the electric current flowing through it. This is H I2 (ii)directly proportional to the resistance of the conductor or resistor. That is, H R (iii)directly proportional to the time for which the electric current flows through the cond...
Read More →Solve this
Question: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when $y=\frac{e^{\operatorname{ex}} \operatorname{sex} x \log x}{\sqrt{1-2 x}}$ Solution: Let $y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{\sqrt{1-2 x}}$ $\Rightarrow y=\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{(1-2 x)^{\frac{1}{2}}}$ Take log both sides: $\Rightarrow \log y=\log \left(\frac{e^{\operatorname{ax}} \sec ^{x} x \log x}{(1-2 x)^{\frac{1}{2}}}\right)$ $\Rightarrow \log y=\log e^{2 x}+\log \sec ^{x} x+\log \log x-\log (1-2 ...
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: $1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots .+$ Solution: To Prove: $\frac{1}{1}+\frac{1}{(1+2)}+\frac{1}{(1+2+3+\ldots \ldots+n)}=\frac{2 n}{(n+1)}$ Let us prove this question by principle of mathematical induction (PMI) Let $P(n): \frac{1}{1}+\frac{1}{(1+2)}+\frac{1}{(1+2+3+\ldots \ldots+n)}=\frac{2 n}{(n+1)}$ For n = 1 $\mathrm{LHS}=1$ $\mathrm{RHS}=\frac{2 \times 1}{(1+1)}=1$ Hence, LHS = RHS ...
Read More →The count of bacteria in a certain experiment was increasing at the rate of 2% per hour.
Question: The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000. Solution: Initial count of bacteria, $P=500000$ Rate of increase, $R=2 \%$ Time, $n=2$ hours Then the count of bacteria at the end of 2 hours is given by Count of bacteria $=P \times\left(1+\frac{R}{100}\right)^{n}$ $=500000 \times\left(1+\frac{2}{100}\right)^{2}$ $=500000 \times\left(\frac{100+2}{100}\right)^{2}$ $=500000 \...
Read More →How will you conclude that the same potential
Question: How will you conclude that the same potential difference (Voltage) exists across three resistors connected in parallel arrangement to a battery ? Solution: Perform an activity to investigate the relation between potential difference across parallel combination of resistors and the potential difference across each individual resistors, Connect three resistors of resistances R1, R2and R3in parallel. One end of each resistor is joined at a common point a and the other end of each resistor...
Read More →The population of a city was 120000 in the year 2009.
Question: The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011? Solution: Population of the city in $2009, \mathrm{P}=120000$ Rate of increase, $\mathrm{R}=6 \%$ Time, $\mathrm{n}=3$ years Then the population of the city in the year 2010 is given by Population $=P \times\left(1+\frac{R}{100}\right)^{n}$ $=120000 \times\left(1+\frac{6}{100}\right)^{1}$ $=12...
Read More →Three years ago, the population of a town was 50000.
Question: Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population? Solution: Let the population of the town be 50000 . Rate of increase for the first year, $p=5 \%$ Rate of increase for the second year, $q=4 \%$ Rate of increase for the third year, $r=3 \%$ Time $=3$ years Now, present population $=\left\{P \times\left(1+\frac{p}{100}\right) \times\left(1+\frac{q}{100}\r...
Read More →How will you infer with the help of an experiment
Question: How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery ? Solution: Perform an activity to show that in series combination of resistors, same current flows through each resistor. Connect three resistors of resistances R1= 1Ω, R2= 2Ω and R3= 3Ω in series. (C.B.S.E. 2014) Connect the series combination of resistors with a battery of 6 V, a plug key K and an ammeter A as s...
Read More →The population of a town is 125000.
Question: The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years? Solution: Population of the town, $P=125000$ Rate of increase, $R=2 \%$ Time, $n=3$ years Then the population of the town after 3 years is given by Population $=P \times\left(1+\frac{R}{100}\right)^{3}$ $=125000 \times\left(1+\frac{2}{100}\right)^{3}$ $=125000 \times\left(\frac{100+2}{100}\right)^{3}$ $=125000 \times\left(\frac{102}{100}\right)^{3}$ $=125000 \tim...
Read More →What is electrical resistivity of a material ?
Question: What is electrical resistivity of a material ? What is its unit ? Describe an experiment to study the factors on which the resistance of conducting wire depends. (CBSE 2012) Solution: For Electrical resistivity:Electrical resistivity of a material is defined as the resistance of an object (made of the material) of unit length and unit area of cross-section. Unit of electrical resistivity is ohm-metre. For experiment: Connect the various electrical components as shown in figure 15. Depe...
Read More →In how many years will Rs 6250 amount to Rs 7290 at 8%
Question: In how many years will Rs 6250 amount to Rs 7290 at 8% per annum, compounded annually? Solution: Let the required time be $n$ years. Rate of interest, $R=8 \%$ Principal amount, $P=$ Rs. 6250 Amount with compound interest, $A=$ Rs. 7290 Then, $A=P \times\left(1+\frac{R}{100}\right)^{n}$ $\Rightarrow$ A $=$ Rs. $6250 \times\left(1+\frac{8}{100}\right)^{n}$ $=$ Rs. $6250 \times\left(\frac{100+8}{100}\right)^{n}$ $=$ Rs. $6250 \times\left(\frac{100}{100}\right)^{n}$ $=$ Rs. $6250 \times\l...
Read More →In how many years will Rs 1800 amount to Rs 2178 at 10%
Question: In how many years will Rs 1800 amount to Rs 2178 at 10% per annum when compounded annually? Solution: Let the required time be $n$ years. Rate of interest, $R=10 \%$ Principal amount, $P=$ Rs. 1800 Amount with compound interest, $A=$ Rs. 2178 Now, $A=P \times\left(1+\frac{R}{100}\right)^{n}$ $=$ Rs. $1800 \times\left(1+\frac{10}{100}\right)^{n}$ $=$ Rs. $1800 \times\left(\frac{100+10}{100}\right)^{\mathrm{n}}$ $=$ Rs. $1800 \times\left(\frac{110}{100}\right)^{\mathrm{n}}$ $=$ Rs. $1800...
Read More →State Ohms law. How can it be verified experimentally ?
Question: State Ohms law. How can it be verified experimentally ? Does it hold good under all conditions ? Comment.(CBSE 2010) Solution: For Ohms law:Ohms law states that the electric current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor, provided the temperature and . other physical conditions of the conductor remain the same.For experimental verification: Verify Ohms lawApparatus :A conductor of resistance R, an ammeter, a vol...
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