Two solutions, A and B, each of

Question: Two solutions, $A$ and $B$, each of $100 \mathrm{~L}$ was made by dissolving $4 \mathrm{~g}$ of $\mathrm{NaOH}$ and $9.8 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in water, respectively. The $\mathrm{pH}$ of the resultant solution obtained from mixing $40 \mathrm{~L}$ of solution $A$ and $10 \mathrm{~L}$ of solution. $B$ is _________________ Solution: (10.60) $\mathrm{M}_{\mathrm{H}_{2} \mathrm{SO}_{4}}=\frac{9.8}{98 \times 100}=10^{-3} \mathrm{M}$ $\mathrm{M}_{\mathrm{NaOH}}=\f...

The value of the integral,

Question: The value of the integral, $\int_{1}^{3}\left[x^{2}-2 x-2\right] \mathrm{d} x$, where $[\mathrm{x}]$ denotes the greatest integer less than or equal to $x$, is:(1) $-4$(2) $-5$(3) $-\sqrt{2}-\sqrt{3}-1$(4) $-\sqrt{2}-\sqrt{3}+1$Correct Option: , 3 Solution: $\mathrm{I}=\int_{1}^{3}-3 d x+\int_{1}^{3}\left[(x-1)^{2}\right] d x$ Put $x-1=t ; d x=d t$' $I=(-6)+\int_{0}^{2}\left[t^{2}\right] d t$ $I=-6+\int_{0}^{1} 0 d t+\int_{1}^{\sqrt{2}} 1 d t+\int_{\sqrt{2}}^{\sqrt{3}} 2 d t+\int_{\sqr...

A 25 m long antenna is mounted on an antenna tower.

Question: A $25 \mathrm{~m}$ long antenna is mounted on an antenna tower. The height of the antenna tower is $75 \mathrm{~m}$. The wavelength (in meter) of the signal transmitted by this antenna would be :(1) 300(2) 400(3) 200(4) 100Correct Option: , 4 Solution: (4) Length of Antena $=25 \mathrm{~m}=\frac{\lambda}{4}$ $\Rightarrow \lambda=100 \mathrm{~m}$...

A 25 m long antenna is mounted on an antenna tower.

Question: A $25 \mathrm{~m}$ long antenna is mounted on an antenna tower. The height of the antenna tower is $75 \mathrm{~m}$. The wavelength (in meter) of the signal transmitted by this antenna would be :(1) 300(2) 400(3) 200(4) 100Correct Option: , 4 Solution: (4)...

A 25 m long antenna is mounted on an antenna tower.

Question: A $25 \mathrm{~m}$ long antenna is mounted on an antenna tower. The height of the antenna tower is $75 \mathrm{~m}$. The wavelength (in meter) of the signal transmitted by this antenna would be :(1) 300(2) 400(3) 200(4) 100Correct Option: , 4 Solution: (4)...

The value of

Question: The value of $\mathrm{Kc}$ is 64 at $800 \mathrm{~K}$ for the reaction $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$. The value of $K_{\mathrm{c}}$ for the following reaction is :$1 / 64$8$1 / 4$$1 / 8$Correct Option: Solution: $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; K_{c}$ $2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3...

If

Question: If $\int_{-\mathrm{a}}^{\mathrm{a}}(|\mathrm{x}|+|\mathrm{x}-2|) \mathrm{dx}=22,(\mathrm{a}2)$ and $[\mathrm{x}]$ denotes the greatest integer $\leq \mathrm{x}$, then $\int_{\mathrm{a}}^{-\mathrm{a}}(\mathrm{x}+[\mathrm{x}]) \mathrm{dx}$ is equal to Solution: $\int_{-a}^{0}(-2 x+2) d x+\int_{0}^{2}(x+2-x) d x+\int_{2}^{a}(2 x-2) d x=22$ $x^{2}-\left.2 x\right|_{0} ^{-a}+\left.2 x\right|_{0} ^{2}+x^{2}-\left.2 x\right|_{2} ^{a}=22$ $a^{2}+2 a+4+a^{2}-2 a-(4-4)=22$ $2 a^{2}=18 \Rightarro...

For the reaction

Question: For the reaction $\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$$K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})$$K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})^{\frac{-1}{2^{\prime}}}$$K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})^{\frac{1}{2}}$$K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})^{\frac{3}{2}}$Correct Option: , 3 Solution: $K_{\mathrm{p}}=K_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm...

Prove the following Definite Integration

Question: $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) d t}{x^{3}}$ is equal to:(1) $\frac{2}{3}$(2) 0(3) $\frac{1}{15}$(4) $\frac{3}{2}$Correct Option: 1 Solution: $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}$ $=\lim _{x \rightarrow 0} \frac{(\sin |x|) 2 x}{3 x^{2}}$ $=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right) \times \frac{2}{3}$ $=\frac{2}{3}$...

Find the HCF of the smallest prime number and the smallest composite number.

Question: Find the HCF of the smallest prime number and the smallest composite number. Solution: Smallest prime number is 2.Smallest composite number is 4.HCF (2, 4) = 2Hence, the HCF of the smallest prime number and the smallest composite number is 2....

Prove the following Definite Integration

Question: $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) d t}{x^{3}}$ is equal to:(1) $\frac{2}{3}$(2) 0(3) $\frac{1}{15}$(4) $\frac{3}{2}$Correct Option: 1 Solution: $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}$...

By using Euclid's algorithm, find the largest number which divides 650 and 1170.

Question: By using Euclid's algorithm, find the largest number which divides 650 and 1170. Solution: 650 and 1170650 1170Thus, we divide 1170 by 650 by using Euclid's division lemma1170 = 650 1 + 520∵Remainder is not zero,we divide 650 by 520 by using Euclid's division lemma650 = 520 1 + 130∵Remainder is not zero,we divide 520 by 130 by using Euclid's division lemma520 = 130 4 + 0Since, Remainder is zero,Therefore, HCF of650 and 1170is 130.Hence, the largest number which divides 650 and 1170 is ...

Lattice enthalpy and enthalpy of solution of

Question: Lattice enthalpy and enthalpy of solution of $\mathrm{NaCl}$ are $788 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $4 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively. The hydration enthalpy of $\mathrm{NaCl}$ is :$-780 \mathrm{~kJ} \mathrm{~mol}^{-1}$$780 \mathrm{~kJ} \mathrm{~mol}^{-1}$$-784 \mathrm{~kJ} \mathrm{~mol}^{-1}$$784 \mathrm{~kJ} \mathrm{~mol}^{-1}$Correct Option: , 3 Solution: $\Delta_{\text {sol. }} \mathrm{H}^{\circ}=\Delta_{\text {lattice }} \mathrm{H}^{\circ}+\Delta_{\text {Hyd...

Using Euclid's division algorithm, find the HCF of

Question: Using Euclid's division algorithm, find the HCF of(i) 612 and 1314(ii) 1260 and 7344(iii) 4052 and 12576 Solution: (i) 612 and 1314612 1314Thus, we divide 1314 by 612 by using Euclid's division lemma1314 = 612 2 + 90∵ Remainder is not zero, we divide 612 by 90by using Euclid's division lemma612 = 90 6 + 72∵Remainder is not zero,we divide 90 by 72 by using Euclid's division lemma90 = 72 1 + 18∵Remainder is not zero,we divide 72 by 18 by using Euclid's division lemma72 = 18 4 + 0Since, R...

Let g(x)

Question: Let $\mathrm{g}(\mathrm{x})=\int_{0}^{x} f(\mathrm{t}) \mathrm{dt}$, where $f$ is continuous function in $[0,3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in[0,1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in(1,3]$ The largest possible interval in which $\mathrm{g}(3)$ lies is:(1) $\left[-1,-\frac{1}{2}\right]$(2) $\left[-\frac{3}{2},-1\right]$(3) $\left[\frac{1}{3}, 2\right]$(4) $[1,3]$Correct Option: , 3 Solution: $\frac{1}{3} \leq f(\mathrm{t}) \leq 1 \forall \mathrm...

Consider the following reaction :

Question: Consider the following reaction : $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}^{0}=+58 \mathrm{~kJ}$ For each of the following cases ((i), (ii)), the direction in which the equilibrium shifts is : (i) Temperature is decreases (ii) Pressure is increased by adding $\mathrm{N}_{2}$ at constant $\mathrm{T}$.(i) towards product, (ii) towards product(i) towards reactant, (ii) towards product(i) towards reactant, (ii) no cha...

Let g(x)

Question: Let $\mathrm{g}(\mathrm{x})=\int_{0}^{x} f(\mathrm{t}) \mathrm{dt}$, where $f$ is continuous function in $[0,3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in[0,1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in(1,3]$ The largest possible interval in which $\mathrm{g}(3)$ lies is:(1) $\left[-1,-\frac{1}{2}\right]$(2) $\left[-\frac{3}{2},-1\right]$(3) $\left[\frac{1}{3}, 2\right]$(4) $[1,3]$Correct Option: , 3 Solution: $\frac{1}{3} \leq f(\mathrm{t}) \leq 1 \forall \mathrm...

Let f(x) and g(x) be two functions satisfying

Question: Let $f(x)$ and $g(x)$ be two functions satisfying $f\left(x^{2}\right)+g(4-x)=4 x^{3}$ and $g(4-x)+g(x)=0$, then the value of $\int_{-4}^{4} f(x)^{2} d x$ is Solution: $\mathrm{I}=2 \int_{0}^{4} f\left(\mathrm{x}^{2}\right) \mathrm{dx}\{$ Even funtion $\}$ $=2 \int_{0}^{4}\left(4 \mathrm{x}^{3}-\mathrm{g}(4-\mathrm{x})\right) \mathrm{d} \mathrm{x}$ $=2\left(\left.\frac{4 \mathrm{x}^{4}}{4}\right|_{0} ^{4}-\int_{0}^{4} \mathrm{~g}(4-\mathrm{x}) \mathrm{d} \mathrm{x}\right)$ $=2(256-0)=5...

The position vector of the centre of mass

Question: The position vector of the centre of mass $\mathrm{r}_{\mathrm{cm}}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is: (1) Correct Option: 1, Solution: $X_{\mathrm{cm}}=\frac{2 \mathrm{~mL}+2 \mathrm{~mL}+\frac{5 \mathrm{~mL}}{2}}{4 \mathrm{~m}}=\frac{13}{8} \mathrm{~L}$ $y$-coordinate of centre of mass is $Y_{c m}=\frac{2 m \times L+m \times\left(\frac{L}{2}\right)+m \times 0}{4 m}=\frac{5 L}{8}$...

Solve the following

Question: If the equilibrium constant for $\mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}$ is $\mathrm{K}_{\mathrm{eq}}^{(1)}$ and that of $\mathrm{B}+\mathrm{C} \rightleftharpoons \mathrm{P}$ is $\mathrm{K}_{\mathrm{eq}}^{(2)}$, the equilibrium constant for $\mathrm{A} \rightleftharpoons \mathrm{P}$ is :$\mathrm{K}_{\mathrm{eq}}^{(1)} / \mathrm{K}_{\mathrm{eq}}^{(2)}$$\mathrm{K}_{\mathrm{eq}}^{(2)}-\mathrm{K}_{\mathrm{eq}}^{(1)}$$\mathrm{K}_{\mathrm{eq}}^{(1)}+\mathrm{K}_{\mathrm{eq}}^{(2)...

An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass.

Question: An alpha-particle of mass $m$ suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, $64 \%$ of its initial kinetic energy. The mass of the nucleus is :(1) $2 \mathrm{~m}$(2) $3.5 \mathrm{~m}$(3) $1.5 \mathrm{~m}$(4) $4 \mathrm{~m}$Correct Option: , 4 Solution: (4) Using conservation of momentum, $\mathrm{mv}_{0}=\mathrm{mv}_{2}-\mathrm{mv}_{1}$ $\frac{1}{2} \mathrm{mv}_{1}^{2}=0.36 \times \frac{1}{2} \mathrm{mv}_{0}^{...

A particle of mass

Question: A particle of mass $20 \mathrm{~g}$ is released with an initial velocity $5 \mathrm{~m} / \mathrm{s}$ along the curve from the point $\mathrm{A}$, as shown in the figure. The point $\mathrm{A}$ is at height $\mathrm{h}$ from point $\mathrm{B}$. The particle slides along the frictionless surface. When the particle reaches point $\mathrm{B}$, its angular momentum about $\mathrm{O}$ will be : (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ) (1) $2 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}$(2) $...

Question: A particle of mass $20 \mathrm{~g}$ is released with an initial velocity $5 \mathrm{~m} / \mathrm{s}$ along the curve from the point $\mathrm{A}$, as shown in the figure. The point $\mathrm{A}$ is at height $\mathrm{h}$ from point $\mathrm{B}$. The particle slides along the frictionless surface. When the particle reaches point $\mathrm{B}$, its angular momentum about $\mathrm{O}$ will be : (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ) (1) $2 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}$(2) $...

A piece of wood of mass

Question: A piece of wood of mass $6.03 \mathrm{~kg}$ is dropped from the top of a $100 \mathrm{~m}$ height building. At the same time, a bullet of mass $0.02 \mathrm{~kg}$ is fired vertically upward, with a velocity $100 \mathrm{~ms}^{-1}$, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$(1) $20 \mathrm{~m}$(2) $30 \mathrm{~m}$(...

The average S-F bond energy in

Question: The average S-F bond energy in $\mathrm{kJmol}^{-1}$ of $\mathrm{SF}_{6}$ is (Rounded off to the nearest integer)[Given : The values of standard enthalpy of formation of $\mathrm{SF}_{6}(\mathrm{~g}), \mathrm{S}(\mathrm{g})$ and $\mathrm{F}(\mathrm{g})$ are $-1100,275$ and $80 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.] Solution: $(309)$ $\mathrm{SF}_{6}(\mathrm{~g}) \longrightarrow \mathrm{S}(\mathrm{g})+6 \mathrm{~F}(\mathrm{~g})$ $\Delta \mathrm{H}_{\text {reaction }}^{\circ}=6 ...