Using prime factorization, find the HCF and LCM of:
Question: Using prime factorization, find the HCF and LCM of:(i) 8, 9, 25(ii) 12, 15, 21(iii) 17, 23, 29(iv) 24, 36, 40(v) 30, 72, 432(vi) 21, 28, 36 Solution: (i) 8, 9, 25Prime factorisation:8 = 2 2 29 = 3 325 = 5 5HCF (8, 9, 25) = 1LCM (8, 9, 25) = 2 2 2 3 3 5 5= 1800(ii) 12, 15, 21Prime factorisation:12 = 2 2 315 = 3 521 = 3 7HCF (12, 15, 21) = 3LCM (12, 15, 21) = 2 2 3 5 7= 420(iii) 17, 23, 29Prime factorisation:17 = 1723 = 2329 = 29HCF (17, 23, 29) = 1LCM (17, 23, 29) = 17 23 29= 11339(iv) ...
Read More →For the equilibrium
Question: For the equilibrium$2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} ;$the value of $\Delta \mathrm{G}^{\circ}$ at $298 \mathrm{~K}$ is approximately:$100 \mathrm{~kJ} \mathrm{~mol}^{-1}$$-80 \mathrm{~kJ} \mathrm{~mol}^{-1}$$80 \mathrm{~kJ} \mathrm{~mol}^{-1}$$-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$Correct Option: , 3 Solution:...
Read More →The integral
Question: The integral $\int_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$ is equal to: (1) $\frac{7}{18}$(2) $-\frac{1}{9}$(3) $-\frac{1}{18}$(4) $\frac{9}{2}$Correct Option: , 3 Solution: $\int_{\pi / 6}^{\pi / 3}\left[\frac{1}{2} \frac{d\left(\tan ^{4} x\right)}{d x} \cdot \sin ^{4} 3 x+\frac{1}{2} \tan ^{4} x \cdot \frac{d\left(\sin ^{4} 3 x\right)}{d x}\right] d x$ $=\frac{1}{2} \int_{\pi / 6}^{\pi / 3} d\left(...
Read More →Let f(x)=|x-2| and g(x)=f(f(x))
Question: Let $f(x)=|x-2|$ and $g(x)=f(f(x)), x \in[0,4]$. Then $\int_{0}^{3}(g(x)-f(x)) d x$ is equal to :(1) 1(2) 0(3) $\frac{1}{2}$(4) $\frac{3}{2}$Correct Option: 1 Solution: $f(x)=|x-2|= \begin{cases}2-x, x2 \\ x-2, x \geq 2\end{cases}$ $g(x)=f(f(x))= \begin{cases}2-f(x), f(x)2 \\ f(x)-2, f(x) \geq 2\end{cases}$ $=\left\{\begin{array}{lll}2-(2-x), 2-x2, x2 \\ (2-x)-2, 2-x \geq 2, x2 \\ 2-(x-2), x-22, x \geq 2 \\ (x-2)-2, x-2 \geq 2, x \geq 2\end{array}\right.$ $=\left\{\begin{array}{cc}-x 0...
Read More →An audio signal
Question: An audio signal $\mathrm{v}_{\mathrm{m}}=20 \sin 2 \pi(1500 \mathrm{t})$ amplitude modulates a carrier $\mathrm{v}_{\mathrm{c}}=80 \sin 2 \pi(100,000 \mathrm{t})$. The value of percent modulation is Solution: Given : $v_{m}=20 \sin \left[100 \pi t+\frac{\pi}{4}\right]$ $v_{\mathrm{C}}=80 \sin \left[10^{4} \pi \mathrm{t}+\frac{\pi}{6}\right]$ we know that, modulation index $=\frac{A_{m}}{A_{c}}$ from given equations, $A_{m}=20$ and $A_{c}=80$ percentage modulation index $=\frac{A_{m}}{A...
Read More →Consider the reaction
Question: Consider the reaction $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$ The equilibrium constant of the above reaction is $\mathrm{K}_{\mathrm{P}}$.If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $\mathrm{P}_{\mathrm{NH}_{3}}\mathrm{P}_{\text {total }}$ at equilibrium $)$ $\frac{3^{3 / 2} K_{P}^{1 / 2} P^{2}}{16}$$\frac{\mathrm{K}_{\mathrm{P}}^{1 / 2} \mathrm{P}^{2...
Read More →Using prime factorization, find the HCF and LCM of
Question: Using prime factorization, find the HCF and LCM of(i) 36, 84(ii) 23, 31(iii) 96, 404(iv) 144, 198(v) 396, 1080(vi) 1152, 1664 Solution: (i) 36, 84Prime factorisation: $36=2^{2} \times 3^{2}$ $84=2^{2} \times 3 \times 7$ $\mathrm{HCF}=$ product of smallest power of each common prime factor in the numbers $=2^{2} \times 3=12$ $\mathrm{LCM}=$ product of greatest power of each prime factor involved in the numbers $=2^{2} \times 3^{2} \times 7=252$ (ii) 23, 31Prime factorisation:23 = 2331 =...
Read More →A TV transmission tower antenna is at a height
Question: A TV transmission tower antenna is at a height of $20 \mathrm{~m}$. Suppose that the receiving antenna is at. (i) ground level (ii) a height of $5 \mathrm{~m}$. The increase in antenna range in case (ii) relative to case (i) is $\mathrm{n} \%$ The value of $\mathrm{n}$, to the nearest integer, is. Solution: (50) Range $=\sqrt{2 \mathrm{Rh}}$ Range (i) $=\sqrt{2 \mathrm{Rh}}$ Range (ii) $=\sqrt{2 \mathrm{Rh}}+\sqrt{2 \mathrm{Rh}^{\prime}}$ where $\mathrm{h}=20 \mathrm{~m} \ \mathrm{~h}^...
Read More →If the value of the integral
Question: If the value of the integral $\int_{0}^{1 / 2} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$ is $\frac{k}{6}$, then $k$ is equal to : (1) $2 \sqrt{3}-\pi$(2) $2 \sqrt{3}+\pi$(3) $3 \sqrt{2}+\pi$(4) $3 \sqrt{2}-\pi$Correct Option: 1 Solution: $\frac{k}{6}=\int_{0}^{\frac{1}{2}} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$ Let $x=\sin \theta ; d x=\cos \theta d \theta$, then $\int_{0}^{\frac{1}{2}} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x=\int_{0}^{\frac{\pi}{6}} \frac{\sin ^{...
Read More →If the value of the integral
Question: If the value of the integral $\int_{0}^{1 / 2} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$ is $\frac{k}{6}$, then $k$ is equal to : (1) $2 \sqrt{3}-\pi$(2) $2 \sqrt{3}+\pi$(3) $3 \sqrt{2}+\pi$(4) $3 \sqrt{2}-\pi$Correct Option: 1 Solution: $\frac{k}{6}=\int_{0}^{\frac{1}{2}} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$ Let $x=\sin \theta ; d x=\cos \theta d \theta$, then $\int_{0}^{\frac{1}{2}} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x=\int_{0}^{\frac{\pi}{6}} \frac{\sin ^{...
Read More →Solve the following
Question: $5.1 \mathrm{~g} \mathrm{NH}_{4} \mathrm{SH}$ is introduced in $3.0 \mathrm{~L}$ evacuated flask at $327^{\circ} \mathrm{C}, 30 \%$ of the solid $\mathrm{NH}_{4} \mathrm{SH}$ decomposed to $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{~S}$ as gases. The $K_{\mathrm{p}}$ of the reaction at $327^{\circ} \mathrm{C}$ is $(\mathrm{R}=0.082$ $\mathrm{L} \mathrm{atm} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$, molar mass of $\mathrm{S}=32 \mathrm{~g} \mathrm{~mol}^{-1}$, molar mass of $\mathrm{N}=14...
Read More →d x is equal to :
Question: $\int_{-\pi}^{\pi}|\pi-| x|| d x$ is equal to : (1) $\sqrt{2} \pi^{2}$(2) $2 \pi^{2}$(3) $\pi^{2}$(4) $\frac{\pi^{2}}{2}$Correct Option: , 3 Solution: $I=\int_{-\pi}^{\pi}|\pi-| x|| d x \quad[\because|\pi-| x||$ is even $]$ $=2 \int_{0}^{\pi}|\pi-| x|| d x$ $=2 \int_{0}^{\pi}(\pi-x) d x$ $=2\left[\pi x-\frac{x^{2}}{2}\right]_{0}^{\pi}=2\left(\pi^{2}-\frac{\pi^{2}}{2}\right)=\pi^{2}$...
Read More →Match List-I with List-II.
Question: Match List-I with List-II. List-I (a) $10 \mathrm{~km}$ height over earth's surface (b) $70 \mathrm{~km}$ height over earth's surface (c) $180 \mathrm{~km}$ height over earth's surface (d) $270 \mathrm{~km}$ height over earth's surface List-II (i) Thermosphere (ii) Mesosphere (iii) Stratosphere (iv) Troposphere(1) (a)-(i v),(b)-(i i i),(c)-(i i),(d)-(i)(2) $(\mathrm{a})-(\mathrm{i}),(\mathrm{b})-(\mathrm{iv}),(\mathrm{c})-(\mathrm{iii}),(\mathrm{d})-(\mathrm{ii})$(3) (a)-(iii), (b)-(ii...
Read More →The integral
Question: The integral $\int_{0}^{2}|| x-1|-x| d x$ is equal to Solution: $\int_{0}^{2}|| x-1|-x| d x=\int_{0}^{1}|1-x-x| d x+\int_{1}^{2}|| x-1-x \mid d x$ $=\int_{0}^{1}(1-2 x) d x+\int_{1 / 2}^{1}(2 x-1) d x+\int_{1}^{2} d x$ $=\left[x-x^{2}\right]_{0}^{\frac{1}{2}}+\left[x^{2}-x\right]_{\frac{1}{2}}^{1}+[x]_{1}^{2}$ $=\frac{1}{2}-\frac{1}{4}+(1-1)-\left(\frac{1}{4}-\frac{1}{2}\right)+2-1=\frac{1}{4}+\frac{1}{4}+1=\frac{3}{2}$...
Read More →in
Question: In $I_{m, n}=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$, for $\mathrm{m}, \mathrm{n} \geq 1$ and $\int_{0}^{1} \frac{\mathrm{x}^{\prime \prime-1}+\mathrm{x}^{n-1}}{(1+\mathrm{x})^{m+n}} \mathrm{dx}=\alpha \mathrm{I}_{\mathrm{m}, \mathrm{n}}, \alpha \in \mathrm{R}$, then $\alpha$ equals Solution: $I_{m, n}=\int_{0}^{1} x^{m-1} \cdot(1-x)^{n-1} d x$ Put $x=\frac{1}{y+1} \Rightarrow d x=\frac{-1}{(y+1)^{2}} d y$ $1-x=\frac{y}{y+1}$ $\therefore I_{m, n}=\int_{\infty}^{0} \frac{y^{n-1}}{(y+1)^{m+...
Read More →The values of
Question: The values of $\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{c}}$ for the following reactions at 300 $\mathrm{K}$ are, respectively: (At $300 \mathrm{~K}, \mathrm{RT}=24.62 \mathrm{dm}^{3} \mathrm{~atm}$ $\mathrm{mol}^{-1}$ ) $1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$$606.0 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$$1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$,$1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$$1,4.1 \time...
Read More →A carrier signal
Question: A carrier signal $\mathrm{C}(\mathrm{t})=25 \sin \left(2.512 \times 10^{10} \mathrm{t}\right)$ is amplitude modulated by a message signal $\mathrm{m}(\mathrm{t})=5 \sin \left(1.57 \times 10^{8} \mathrm{t}\right)$ and transmitted through an antenna. What will be the bandwidth of the modulated signal?(1) $8 \mathrm{GHz}$(2) $2.01 \mathrm{GHz}$(3) $1987.5 \mathrm{MHz}$(4) $50 \mathrm{MHz}$Correct Option: , 4 Solution: (4) Band width $=2 \mathrm{f}_{\mathrm{m}}$ $\omega_{\mathrm{m}}=1.57 \...
Read More →Which one of the following will be the output of the given circuit?
Question: Which one of the following will be the output of the given circuit? (1) NOR Gate(2) AND Gate(3) AND Gate(4) XOR GateCorrect Option: , 4 Solution: (4) Conceptual...
Read More →Let f(x) =
Question: Let $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x}$ be a differentiable function for all $x \in R$. Then $\mathrm{f}(\mathrm{x})$ equals.(1) $2 e^{\left(e^{2}-1\right)}-1$(2) $\mathrm{e}^{\left(\mathrm{e}^{x}-1\right)}$(3) $2 e^{e^{3}}-1$(4) $e^{\tau^{x}}-1$Correct Option: 1 Solution: Given, $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x} \ldots(1)$ Differentiating both sides w.r.t $x f^{\prime}(x)=e^{x} \cdot f(x)+e^{x}$ (Using Newton Leibnitz Theorem) $\Rightarrow \frac{f^{\prime}(z)}{f(x)+1}=e^{x}...
Read More →Express 2431 as a product of its prime factors.
Question: Express 2431 as a product of its prime factors. Solution: Prime Factorisation of 2431 is:2431 = 11 13 17Hence, 2431 as a product of its prime factors can be expressed as 11 13 17....
Read More →The HCF of two numbers is 145 and their LCM is 2175.
Question: The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other. Solution: HCF of two numbers = 145LCM of two numbers = 2175Let one of the two numbers be 725and other bex.Using the formula, Product of two numbers = HCF LCMwe conclude that 725x= 145 2175 $x=\frac{145 \times 2175}{725}$ = 435 Hence, the other number is 435....
Read More →For VHF signal broadcasting,
Question: For VHF signal broadcasting, $\mathrm{km}^{2}$ of maximum service area will be covered by an antenna tower of height $30 \mathrm{~m}$, if the receiving antenna is placed at ground. Let radius of the earth be $6400 \mathrm{~km}$. (Round off to the Nearest Integer) (Take $\pi$ as $3.14$ ) Solution: $(1206)$ $d=\sqrt{2 R h}$ $\mathrm{~A}=\pi \mathrm{d}^{2}$ $\mathrm{~A}=\pi 2 \mathrm{Rh}$ $=3.14 \times 2 \times 6400 \times \frac{30}{1000}$ $\mathrm{A}=1205.76 \mathrm{~km}^{2}$ $\mathrm{~A...
Read More →The HCF of two numbers is 23 and their LCM is 1449.
Question: The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other. Solution: Let the two numbers beaandb.Let the value ofabe161.Given:HCF = 23 and LCM = 1449we know, ab=HCF LCM⇒ 161 b = 23 1449 $\Rightarrow \quad \therefore b=\frac{23 \times 1449}{161}=\frac{33327}{161}=207$ Hence, the other number bis 207....
Read More →For x>0
Question: For $x0$, if $f(x)=\int_{1}^{x} \frac{\log _{e} t}{(1+t)} d t$, then $f(e)+f\left(\frac{1}{\epsilon}\right)$ is equal to :(1) $\frac{1}{2}$(2) $-1$(3) 1(4) 0Correct Option: 1 Solution: $f(e)+f\left(\frac{1}{e}\right)=\int_{1}^{e} \frac{\ell n t}{1+t} d t+\int_{1}^{1 / e} \frac{\ell n t}{1+t} d t=I_{1}+I_{2}$ $I_{2}=\int_{1}^{1 / e} \frac{\text { ent }}{1+t} d t \quad$ put $t=\frac{1}{z}, d t=-\frac{d z}{z^{2}}$ $=\int_{1}^{e}-\frac{\ell n z}{1+\frac{1}{3}} \times\left(-\frac{d z}{z^{2}...
Read More →Consider the following reversible chemical reactions:
Question: Consider the following reversible chemical reactions: $\mathrm{K}_{1} \mathrm{~K}_{2}=\frac{1}{3}$$\mathrm{K}_{2}=\mathrm{K}_{1}{ }^{3}$$\mathrm{K}_{2}=\mathrm{K}_{1}^{-3}$$\mathrm{K}_{1} \mathrm{~K}_{2}=3$Correct Option: , 3 Solution:...
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