For the equilibrium
Question: For the equilibrium $\mathrm{A} \rightleftharpoons \mathrm{B}$, the variation of the rate of the forward (a) and reverse (b) reaction with time is given by:Correct Option: , 2 Solution: At equilibrium, rate of forward reaction = Rate of backward reaction....
Read More →Let f be a differentiable function from
Question: Let $f$ be a differentiable function from $\mathbf{R}$ to $\mathbf{R}$ such that $|f(x)-f(y)| \leq 2|x-y|^{3 / 2}$, for all $x, y, \in \mathbf{R}$. If $f(0)=1$ then $\int_{0}^{1} f^{2}(x) d x$ is equal to :(1) 1(2) 2(3) $\frac{1}{2}$(4) 0Correct Option: 1 Solution: $\because f: R \rightarrow R$ and $|f(x)-f(y)| \leq 2 \cdot|x-y|^{3 / 2}$ $\Rightarrow \quad\left|\frac{f(x) \quad f(y)}{x \quad y}\right| \leq \sqrt{x-y}$ $\Rightarrow \quad \lim _{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y...
Read More →An electronic device makes a beep after every 60 seconds. Another device makes a beep after 62 seconds.
Question: An electronic device makes a beep after every 60 seconds. Another device makes a beep after 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest? Solution: Beep duration of first device = 60 secondsBeep duration of second device = 62 seconds Interval of beeping together = LCM(60, 62)Prime factorisation: $60=2^{2} \times 3 \times 5$ $62=2 \times 31$ $\therefore \mathrm{LCM}=2^{2} \times 3 \times 5 \times 31=1860$ seconds $=\frac{1860}{60}=31 \...
Read More →The value of
Question: The value of $\int_{0}^{\pi}|\cos x|^{3} \mathrm{~d} x$ is:(1) 0(2) $\frac{4}{3}$(3) $\frac{2}{3}$(4) $\frac{-4}{3}$Correct Option: , 2 Solution: $I=\int_{0}^{\pi}|\cos x|^{3} d x$ $=2 \int_{0}^{\pi / 2} \cos ^{3} x d x$ $=\frac{2}{4} \int_{0}^{\pi / 2}(3 \cos x+\cos 3 x) d x$ $\left[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right]$ $=\frac{1}{2}\left[3 \sin x+\frac{\sin 3 x}{3}\right]_{0}^{\pi / 2}$ $=\frac{1}{2}\left(3-\frac{1}{3}\right)=\frac{4}{3}$...
Read More →Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times,
Question: Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods. Solution: Length of the three measuring rods are 64 cm, 80 cm and 96 cm, respectively. Length of cloth that can be measured an exact number of times = LCM (64, 80, 96)Prime factorisation: $64=2^{6}$ $80=2^{4} \times 5$ $96=2^{5} \times 3$ $\therefore$ LCM $=$ Product of greatest power of each prime factor involved in the numbers...
Read More →For the reaction
Question: For the reaction $2 \mathrm{~A}+3 \mathrm{~B}+\frac{3}{2} \mathrm{C} \rightarrow 3 \mathrm{P}$, which statement is correct?$\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{3}{2} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{3}{4} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}$$\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}$$\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm...
Read More →A value of $lpha$ such that
Question: A value of $\alpha$ such that $\int_{\alpha}^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}=\log _{e}\left(\frac{9}{8}\right)$ is :(1) $-2$(2) $\frac{1}{2}$(3) $-\frac{1}{2}$(4) 2Correct Option: 1 Solution: $\int_{\alpha}^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}$ $=\int_{\alpha}^{\alpha+1}\left[\frac{1}{x+\alpha}-\frac{1}{x+\alpha+1}\right] d x$ [Using partial fraction] $=\left.\log \left(\frac{(x+\alpha)}{(x+\alpha+1)}\right)\right|_{\alpha} ^{\alpha+1}=\log \left(\frac{2 \alpha+1...
Read More →Find the leash number of square tiles required to pave the ceiling of a room 15 m 17
Question: Find the leash number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad. Solution: It is given that: Length of a tile = 15 m 17 cm = 1517 cm [∵1 m = 100 cm]Breadth of a tile = 9 m 2 cm = 902 cmSide of each square tile = HCF (1517, 902)Prime factorisation: $1517=37 \times 41$ $902=22 \times 41$ HCF = Product of smallest power of each common prime factor in the numbers = 41 $\therefore$ Required number of tiles $=\frac{\text { Area of ceiling }}{\t...
Read More →It is true that:
Question: It is true that:A second order reaction is always a multistep reactionA zero order reaction is a multistep reactionA first order reaction is always a single step reactionA zero order reaction is a single step reactionCorrect Option: , 2 Solution: Zero order reaction is always multi step reaction....
Read More →If
Question: If $\int_{0}^{\frac{x}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$, then $m \cdot n$ is equal to :(1) $-\frac{1}{2}$(2) 1(3) $\frac{1}{2}$(4) $-1$Correct Option: , 4 Solution: $\int_{0}^{\pi / 2} \frac{\cot x d x}{\cot x+\operatorname{cosec} x}$ $=\int_{0}^{\pi / 2} \frac{\cot x d x}{1+\cos x}=\int_{0}^{\pi / 2}\left(1-\frac{1}{1+\cos x}\right) d x$ $=[x]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \frac{1}{2 \cos ^{2} \frac{x}{2}} d x=\frac{\pi}{2}-\frac{1}{2} \int_{0}^{\pi / ...
Read More →The results given in the below table were obtained during kinetic studies of the following reaction :
Question: The results given in the below table were obtained during kinetic studies of the following reaction : $0.4,0.4$$0.4,0.3$$0.3,0.4$$0.3,0.3$Correct Option: , 3 Solution: $\operatorname{Rate}(\mathrm{R})=\mathrm{k}[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}$ $\operatorname{Exp} \mathrm{I} \Rightarrow 6.0 \times 10^{-3}=\mathrm{k}[0.1]^{\mathrm{a}}[0.1]^{\mathrm{b}}$ $\operatorname{Exp} \mathrm{II} \Rightarrow 24.0 \times 10^{-3}=\mathrm{k}[0.1]^{\mathrm{a}}[0.2]^{\mathrm{b}}$ $\ope...
Read More →Find the maximum number of students among whom 1001 pens and 910
Question: Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils. Solution: Total number of pens = 1001Total number of pencils = 910Maximum number of students who get the same number of pens and pencils =HCF (1001, 910)Prime factorisation: $1001=11 \times 91$ $910=10 \times 91$ $\therefore \mathrm{HCF}=91$ Hence, 91 students receive same number of pens and pencils....
Read More →Let f:
Question: Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be a continuously differentiable function such that $f(2)=6$ and $f^{\prime}(2)=\frac{1}{48}$. If $\int_{6}^{f(x)} 4 t^{3} d t=(x-2) g(x)$, then $\lim _{x \rightarrow 2} g(x)$ is equal to : (1) 18(2) 24(3) 12(4) 36Correct Option: 1 Solution: Given, $\int_{6}^{f(x)} 4 t^{3} d t=(x-2) g(x)$ Differentiating both sides, $4(f(x))^{3} \cdot f^{\prime}(x)=g^{\prime}(x)(x-2)+g(x)$ Putting $x=2, \frac{4(6)^{3} \cdot 1}{48}=g(2) \Rightarrow \lim _{x \ri...
Read More →Find the greatest possible length which can be used to measure exactly the length 7 m,
Question: Find the greatest possible length which can be used to measure exactly the length 7 m, 3 m 85 cm and 12 m 95 cm. Solution: The three given lengths are 7 m (700 cm), 3 m 85 cm (385 cm) and 12 m 95 cm (1295 cm). (∵ 1 m = 100 cm)Required length = HCF (700, 385, 1295)Prime factorisation: $700=2 \times 2 \times 5 \times 5 \times 7=2^{2} \times 5^{2} \times 7$ $385=5 \times 7 \times 11$ $1295=5 \times 7 \times 37$ $\therefore \mathrm{HCF}=5 \times 7=35$ Hence, the greatest possible length is...
Read More →Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length.
Question: Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank? How many planks are formed? Solution: The lengths of three pieces of timber are42 m, 49 m and63 m, respectively.We have to divide the timber into equal length of planks.Greatest possible length of each plank = HCF(42, 49, 63)Prime factorisation: $42=2 \times 3 \times 7$ $49=7 \times 7$ $63=3 \times 3 \times 7$ HCF = Product of smallest p...
Read More →If the activation energy of a reaction is
Question: If the activation energy of a reaction is $80.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the fraction of molecules at $700 \mathrm{~K}$, having enough energy to react to form products is $\mathrm{e}^{-\mathrm{x}}$. The value of $\mathrm{x}$ is _________________ (Rounded off to the nearest integer) $\left[\mathrm{Use} \mathrm{R}=8.31 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right]$ Solution: (14) $\mathrm{E}_{\mathrm{a}}=80.9 \mathrm{~kJ} / \mathrm{mol}$ Eraction of molecules able to cross energy b...
Read More →Three sets of English, Mathematics and Science books containing 336,
Question: Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the higher of each stack is the same. How many stacks will be there? Solution: Total number of English books = 336Total number of mathematics books = 240Total number of science books = 96Number of books stored in each stack = HCF (336, 240, 96)Prime factorisation: $336=2^{4} \times 3 \times 7$ $240=2^{4} \time...
Read More →An exothermic reaction
Question: An exothermic reaction $X \rightarrow Y$ has an activation energy $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If energy change $\Delta \mathrm{E}$ during the reaction is $-20 \mathrm{~kJ}$, then the activation energy for the reverse reaction in $\mathbf{k J}$ is Solution: (50) $\Delta \mathrm{H}=\mathrm{E}_{\mathrm{a}, \mathrm{f}}-\mathrm{E}_{\mathrm{a}, \mathrm{b}}$ $-20=30-\mathrm{E}_{\mathrm{a}, \mathrm{b}}$ $\mathrm{E}_{\mathrm{a}, \mathrm{b}}=50 \mathrm{~kJ} / \mathrm{mole}$...
Read More →In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively.
Question: In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject. Solution: Minimum number of rooms required $=\frac{\text { Total number of participants }}{\operatorname{HCF}(60,84,108)}$ Prime factorization of 60, 84 and 108 is: $60=2^{2} \times 3 \times 5$ $84=2^{2} \times 3 \times 7$ $108=2^...
Read More →The rate constant of a reaction increases by five times on increase in temperature from
Question: The rate constant of a reaction increases by five times on increase in temperature from $27^{\circ} \mathrm{C}$ to $52^{\circ} \mathrm{C}$. The value of activation energy in $\mathrm{kJmol}^{-1}$ is ___________(Rounded off to the nearest integer) $\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$ Solution: (52)...
Read More →Find the least number which when divided by 20, 25, 35 and 40 leaves remainders
Question: Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively. Solution: First find the LCM of 20, 25, 35 and 40. $\operatorname{LCM}(20,25,35,40)=2 \times 2 \times 2 \times 5 \times 5 \times 7=1400$ Now, we can see that $20-14=6$ $25-19=6$ $35-29=6$ $40-34=6$ So, the required number would beLCM $(20,25,35,40)-6$ $=1400-6$ $=1394$...
Read More →Solve the following
Question: For the reaction, $a A+b B \rightarrow c C+d D$, the plot of $\log k v s \frac{1}{T}$ is given below: The temperature at which the rate constant of the reaction is $10^{-4} \mathrm{~s}^{-1}$ is K. [Rounded off to the nearest integer) [Given: The rate constant of the reaction is $10^{-5} \mathrm{~s}^{-1}$ at $500 \mathrm{~K}$ ] Solution:...
Read More →Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Question: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. Solution: We need to find the greatest number that would divide 43, 91 and 183 leaving the same remainder every time.We first find the difference of the numbers and then find the HCF of the got numbers. $183-91=92$ $183-43=140$ $91-43=48$ Now find HCF of 92, 140 and 48, we get $92=2 \times 2 \times 23$ $140=2 \times 2 \times 5 \times 7$ $48=2 \times 2 \times 2 \times 2 \times 3$ HCF...
Read More →The integral
Question: The integral $\int_{\pi / 6}^{\pi / 3} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ is equal to :(1) $3^{5 / 6}-3^{2 / 3}$(2) $3^{4 / 3}-3^{1 / 3}$(3) $3^{7 / 6}-3^{5 / 6}$(4) $3^{5 / 3}-3^{1 / 3}$Correct Option: , 3 Solution: Let, $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 \cdot d x}{\cos ^{\frac{2}{3}} x \cdot \sin ^{\frac{4}{3}} x}$ $=\int_{\frac{\pi}{6}}^{\frac{\pi}...
Read More →Find the least number which should be added to 2497
Question: Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3. Solution: We find the LCM of 5, 6, 4 and 3 first. So, $\operatorname{LCM}(5,6,4,3)=2 \times 2 \times 3 \times 5=60$ Now, divide 2497 by 60, we get To make the number completely divisible by 60, we must add a number that would make the remainder equal to 60.Therefore, the number that must be added is 60-37 = 23Hence, 23 must be added to 2497.So, the number exactly divisible by 5, ...
Read More →