Express 5005 as a product of its prime factors.

Question: Express 5005 as a product of its prime factors. Solution: Prime Factorisation of 5005 is:5005 = 5 7 11 13​Hence, 5005 as a product of its prime factors can be expressed as 5 7 11 13....

Read More →

The output of the given combination gates

Question: The output of the given combination gates represents : (1) XOR Gate(2) NAND Gate(3) AND Gate(4) NOR GateCorrect Option: , 2 Solution: (2) By De Morgan's theorem, we have:...

Read More →

Express 429 as a product of its prime factors.

Question: Express 429 as a product of its prime factors. Solution: Prime factorisation of 429 is:429 =3 11 13Hence, 429 as a product of its prime factors can be expressed as 3 11 13....

Read More →

In which one of the following equilibria,

Question: In which one of the following equilibria, $K_{p} \neq K_{c}$ ?$2 \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})$$\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\mathrm{SO}_{3}(\mathrm{~g})$$2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm...

Read More →

Using Euclid's algorithm, find the largest number that divides 1251,

Question: Using Euclid's algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2, and 3 respectively. Solution: On subtracting1, 2, and 3 from1251, 9377 and 15628 respectively, we get 1250, 9375 and 15625.Now we find the HCF of1250 and 9375 using Euclid's division lemma1250 9375Thus, we divide 9375 by 1250 by using Euclid's division lemma9375 = 1250 7 + 625∵ Remainder is not zero, we divide 1250 by 625 by using Euclid's division lemma1250 = 625 2 + 0Since, Re...

Read More →

Use Euclid's algorithm to find the HCF of 441, 567 and 693.

Question: Use Euclid's algorithm to find the HCF of 441, 567 and 693. Solution: Let us first find the HCF of 441 and 567using Euclid's division lemma.441 567Thus, we divide 567 by 441 by using Euclid's division lemma567 = 441 1 + 126∵ Remainder is not zero, we divide 441 by 126 by using Euclid's division lemma441 = 126 3 + 63∵Remainder is not zero,we divide 126 by 63 by using Euclid's division lemma126 = 63 2 + 0Since, Remainder is zero,Therefore, HCF of441 and 567 is 63.Now, let us find the HCF...

Read More →

For the following reactions, equilibrium constants are given :

Question: For the following reactions, equilibrium constants are given : $\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}=10^{52}$ $2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \mathrm{K}_{2}=10^{129}$ The equilibrium constant for the reaction, $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$ is :$10^{154...

Read More →

The value of the integral

Question: The value of the integral $\int_{0}^{\pi}|\sin 2 x| d x$ is Solution: $\mathrm{I}=\int_{0}^{\pi}|\sin 2 \mathrm{x}| \mathrm{d} \mathrm{x}$ $\mathrm{I}=2 \int_{0}^{\pi / 2}|\sin 2 \mathrm{x}| \mathrm{dx}=2 \int_{0}^{\pi / 2} \sin 2 \mathrm{xd} \mathrm{x}$ $\mathrm{I}=2\left[\frac{-\cos (2 \mathrm{x})}{2}\right]_{0}^{\pi / 2}=2$...

Read More →

The value of

Question: The value of $\int_{-\pi / 2}^{\% / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ is:(1) $2 \pi$(2) $4 \pi$(3) $\frac{\pi}{2}$(4) $\frac{\pi}{4}$Correct Option: , 4 Solution: Let $I=\int_{-\pi / 2}^{5 / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ $I=\int_{-\pi / 2}^{\pi / 2} \frac{3^{x} \cos ^{2} x}{1+3^{x}} d x$ $2 I=\int_{-\pi / 2}^{/ 2} \cos ^{2} x d x$ $I=\int_{0}^{\pi / 2} \cos ^{2} x d x=\frac{\pi}{4}$...

Read More →

In the figure shown below reactant A (represented by square) is in equilibrium with product B (represented by circle).

Question: In the figure shown below reactant $A$ (represented by square) is in equilibrium with product B (represented by circle). The equilibrium constant is: 4812Correct Option: , 4 Solution: Equilibrium constant $=\frac{[B]}{[A]} \approx 2$...

Read More →

The value of

Question: The value of $\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} d x$, where $[x]$ is the greatest integer $\leq x$, is:(1) $100(e-1)$(2) $100 e$(3) $100(1-e)$(4) $100(1+\mathrm{e})$Correct Option: 1 Solution: $\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} d x$ $=\int_{0}^{1} e^{[x\}} d x+\int_{1}^{2} e^{\{x\}} d x+\int_{2}^{3} e^{\{x\}} d x+\ldots \ldots \int_{99}^{100} e^{\{x\}} d x \quad(\because\{x\}=x-[x])$ $=\left.e^{x}\right|_{0} ^{1}+\left.e^{(x-1)}\right|_{1} ^{2}+\left.e^{(x-2)}\right|_{2...

Read More →

Use Euclid's algorithm to find HCF of 1190 and 1445.

Question: Use Euclid's algorithm to find HCF of 1190 and 1445. Express the HCF in the form 1190m+ 1445n. Solution: Using Euclid's division algorithm, we have Since 1445 1190, we apply Euclid's division lemma to 1445 and 1190 to get; $1445=1190 \times 1+255$ Since the remainder is not zero, we again apply division lemma to 1190 and 255 and get; $1190=255 \times 4+170$ Again, the remainder is not zero, so we apply division lemma to 255 and 170 to get; $255=170 \times 1+85$ Now we finally apply div...

Read More →

Two identical antennas mounted on identical towers are separated from each other

Question: Two identical antennas mounted on identical towers are separated from each other by a distance of $45 \mathrm{~km}$. What should nearly be the minimum height of receiving antenna to receive the signals in line of sight? (Assume radius of earth is $6400 \mathrm{~km}$ )(1) $19.77 \mathrm{~m}$(2) $39.55 \mathrm{~m}$(3) $79.1 \mathrm{~m}$(4) $158.2 \mathrm{~m}$Correct Option: 2, Solution: (2) $\mathrm{D}=2 \sqrt{2 \mathrm{Rh}}$ $\mathrm{h}=\frac{\mathrm{D}^{2}}{8 \mathrm{R}}=\frac{45^{2}}{...

Read More →

The value of

Question: The value of $\int_{2}^{2}\left|3 x^{2}-3 x-6\right| d x$ is Solution: $3 \int_{-2}^{2}\left|x^{2}-x-2\right| d x x^{2}-x-2$ $=(x-2)(x+1)$ $=3\left\{\int_{-2}^{-1}\left(x^{2}-x-2\right) d x+\int_{-1}^{2}\left(-x^{2}+x+2\right) d x\right\}$ $=3\left[\left(\frac{x^{3}}{3}-\frac{x^{2}}{2}-2 x\right)_{-2}^{-1}-\left(\frac{x^{3}}{3}-\frac{x^{2}}{2}-2 x\right)_{-1}^{2}\right]$ $=19$...

Read More →

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

Question: The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively: $\mathrm{X}_{2} \mathrm{Y}, 2 \times 10^{-9} \mathrm{M}^{3}$$\mathrm{XY}_{2}, 4 \times 10^{-9} \mathrm{M}^{3}$$\mathrm{XY}_{2}, 1 \times 10^{-9} \mathrm{M}^{3}$$X Y, 2 \times 10^{-6} \mathrm{M}^{3}$Correct Option: , 2 Solution: From the given curve, if $[X]=1 \mathrm{mM}$ then $[Y]=2 \mathrm{mM}$ $\therefore \quad$ Salt is $\mathrm{XY}_{2}$ $\mathrm{K}_{\mathrm{sp}}=[\mathrm{X}][\...

Read More →

Prove that if x and y are both odd positive integers then

Question: Prove that if $x$ and $y$ are both odd positive integers then $x^{2}+y^{2}$ is even but not divisible by 4 . Solution: Let, $n$ be any positive odd integer and let $x=n$ and $y=n+2$. So, $x^{2}+y^{2}=(n)^{2}+(n+2)^{2}$ Or, $x^{2}+y^{2}=n^{2}+\left(n^{2}+4+4 n\right)$ $\Rightarrow x^{2}+y^{2}=2 n^{2}+4+4 n$ $\Rightarrow x^{2}+y^{2}=2\left(n^{2}+2+2 n\right)$ $\Rightarrow x^{2}+y^{2}=2 m$ (where $m=n^{2}+2 n+2$ ) Because $x^{2}+y^{2}$ has 2 as a factor, so the value is an even number. Al...

Read More →

The following logic gate is equivalent to :

Question: The following logic gate is equivalent to : (1) NOR Gate(2) OR Gate(3) AND Gate(4) NAND GateCorrect Option: 1 Solution: (1) Truth table for the given logic gate : The truth table is similar to that of a NOR gate....

Read More →

The following logic gate is equivalent to :

Question: The following logic gate is equivalent to : (1) NOR Gate(2) OR Gate(3) AND Gate(4) NAND GateCorrect Option: 1 Solution: (1) Truth table for the given logic gate : The truth table is similar to that of a NOR gate....

Read More →

is equal to:

Question: $\lim _{x \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^{2}}+\frac{n}{(n+2)^{2}}+\ldots \cdots+\frac{n}{(2 n-1)^{2}}\right]$ is equal to:(1) 1(2) $\frac{1}{3}$(3) $\frac{1}{2}$(4) $\frac{1}{4}$Correct Option: , 3 Solution: $\lim _{x \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^{2}}=\operatorname{Lim}_{x \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n^{2}}{n^{2}\left(1+\frac{r}{u}\right)^{2}}=\int_{0}^{1} \frac{d x}{(1+x)^{2}}$ $=-\left[\frac{1}{1+x}\right]_{0}^{1} \Rightarrow...

Read More →

The ammonia

Question: The ammonia $\left(\mathrm{NH}_{3}\right)$ released on quantitative reaction of $0.6 \mathrm{~g}$ urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ with sodium hydroxide $(\mathrm{NaOH})$ can be neutralized by:$200 \mathrm{~mL}$ of $0.4 \mathrm{~N} \mathrm{HCl}$$200 \mathrm{~mL}$ of $0.2 \mathrm{NHCl}$$100 \mathrm{~mL}$ of $0.2 \mathrm{~N} \mathrm{HCl}$$100 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}$Correct Option: , 3 Solution: $1 \mathrm{~mol}$ of urea $=2 \mathrm{~mol}$ of $...

Read More →

If

Question: If $\mathrm{I}_{\mathrm{n}}=\int_{\pi / 4}^{\pi / 2} \cot ^{\mathrm{n}} \mathrm{xdx}$, then:(1) $\frac{1}{\mathrm{I}_{2}+\mathrm{L}_{4}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{5}}, \frac{1}{\mathrm{~L}_{4}+\mathrm{I}_{6}}$ are in G.P.(2) $\frac{1}{\mathrm{I}_{2}+\mathrm{I}_{1}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{6}}, \frac{1}{\mathrm{I}_{1}+\mathrm{I}_{6}}$ are in A.P.(3) $\mathrm{I}_{2}+\mathrm{I}_{4}, \mathrm{I}_{3}+\mathrm{I}_{5}, \mathrm{I}_{4}+\mathrm{I}_{6}$ are in A.P.(4) $\math...

Read More →

For any positive integer n, prove that n3 – n is divisible by 6.

Question: For any positive integern, prove thatn3nis divisible by 6. Solution: Euclid's division lemma states that for given positive integers $a$ and $b$, there exists unique integers $q$ and $r$ satisfying $a=b q+r, 0 \leq rb$ Applying Euclid's division lemma omnand 6, we have $n=6 q+r, 0 \leq r6$ Therefore,n can have six values, i.e. $n=6 q$ $n=6 q+1$ $n=6 q+2$ $n=6 q+3$ $n=6 q+4$ $n=6 q+5$ Case I: When $n=6 q$ $n^{3}=(6 q)^{3}$ $n^{3}-n=(6 q)^{3}-6 q$ $=6 q\left(36 q^{2}-1\right)$ $=6 m\left...

Read More →

In the logic circuit shown in the figure,

Question: In the logic circuit shown in the figure, if input $\mathrm{A}$ and $B$ are 0 to 1 respectively, the output at $Y$ would be ' $x$ '. The value of $x$ is Solution:...

Read More →

The value of

Question: The value of $\int_{-1}^{1} \mathrm{x}^{2} \mathrm{e}^{\left[\mathrm{x}^{3}\right]} \mathrm{dx}$, where $[\mathrm{t}]$ denotes the greatest integer $\leq \mathrm{t}$, is :(1) $\frac{\mathrm{e}+1}{3}$(2) $\frac{e-1}{3 e}$(3) $\frac{\mathrm{e}+1}{3 \mathrm{e}}$(4) $\frac{1}{3 \mathrm{e}}$Correct Option: , 3 Solution: $I=\int_{-1}^{0} x^{2} \cdot e^{-1} d x+\int_{0}^{1} x^{2} d x$ $\therefore I=\left.\frac{x^{3}}{3 e}\right|_{-1} ^{0}+\left.\frac{x^{3}}{3}\right|_{0} ^{1}$ $\Rightarrow I=...

Read More →

Let f(x) be a differentiable function

Question: Let $f(x)$ be a differentiable function defined on $[0,2]$ such that $f^{\prime}(x)=f^{\prime}(2-x)$ for all $x \in(0,2), f(0)=1$ and $f(2)=\mathrm{e}^{2}$. Then the value of $\int_{0}^{2} f(x) \mathrm{d} x$ is:(1) $1+e^{2}$(2) $1-\mathrm{e}^{2}$(3) $2\left(1-e^{2}\right)$(4) $2\left(1+e^{2}\right)$Correct Option: 1 Solution: $f^{\prime}(x)=f^{\prime}(2-x)$ On integrating both side $f(x)=-f(2-x)+c$ put $x=0 f(0)+f(2)=c \quad \Rightarrow c=1+e^{2}$ $\Rightarrow \mathrm{f}(\mathrm{x})+\m...

Read More →