A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder.
Question: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Solution: For hemispherical part, radius (r) $=\frac{\mathbf{1 4}}{\mathbf{2}}=7 \mathrm{~cm}$ $\therefore \quad$ Curved surface area $=2 \pi \mathrm{r}^{2}$ $=2 \times \frac{22}{7} \times 7 \times 7 \mathrm{~cm}^{2}=308 \mathrm{~cm}^{2}$ Total height of vessel $=13 \mathrm{~cm}$ $\th...
Read More →Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions.
Question: Calculate the entropy change in surroundings when $1.00 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}_{(f)}$ is formed under standard conditions. $\Delta \mathrm{f}^{\theta}=-286 \mathrm{~kJ}$ mol $^{-1}$. Solution: It is given that $286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ of heat is evolved on the formation of $1 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}_{(j)}$. Thus, an equal amount of heat will be absorbed by the surroundings. $q_{\text {surr }}=+286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ...
Read More →2 cubes each of volume 64 cm
Question: 2 cubes each of volume $64 \mathrm{~cm}^{3}$ are joined end to end. Find the surface area of the resulting cuboid. Solution: Let $\ell \mathrm{cm}$ be the length of an edge of the cube having volume $=64 \mathrm{~cm}^{3}$. Then, $\ell^{3}=64=(4)^{3} \Rightarrow \ell=4 \mathrm{~cm}$ Now, the dimensions of the resulting cuboid made by joining two cubes (see figure) are 8 cm 4 cm 4 cm (i.e., length = 8 cm, breadth $=4 \mathrm{~cm}$ and height $=4 \mathrm{~cm}$ ) Surface area of cuboid $=2...
Read More →Given an example of a relation. Which is
Question: Given an example of a relation. Which is (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. (iii) Reflexive and symmetric but not transitive. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive. Solution: (i) LetA= {5, 6, 7}. Define a relation R onAas R = {(5, 6), (6, 5)}. Relation R is not reflexive as (5, 5), (6, 6), (7, 7) R. Now, as $(5,6) \in R$ and also $(6,5) \in R, R$ is symmetric....
Read More →Comment on the thermodynamic stability
Question: Comment on the thermodynamic stability of $\mathrm{NO}_{(g) \text {, given }}$ $\frac{1}{2} \mathrm{~N}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{(g)} ; \Delta_{r} H^{\theta}=90 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\mathrm{NO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{2(g)}: \Delta_{r} H^{\theta}=-74 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Solution: The positive value of $\Delta_{r} H$ indicates that heat is absorbed during the formation of $\mathrm{NO}_{(g...
Read More →The equilibrium constant for a reaction is 10. What will be the value
Question: The equilibrium constant for a reaction is $10 .$ What will be the value of $\Delta G^{\theta} ? \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, T=300 \mathrm{~K}$. Solution: From the expression, $\Delta G^{\theta}=-2.303 \mathrm{RT} \log K_{e q}$ $\Delta G^{\theta}$ for the reaction, $=(2.303)\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K}) \log 10$ $=-5744.14 \mathrm{Jmol}^{-1}$ $=-5.744 \mathrm{~kJ} \mathrm{~mol}^{-1}$...
Read More →Calculate the area of the designed region in fig.
Question: Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each. Solution: Side of the square = 8 cm $\therefore$ Area of the square $(\mathrm{ABCD})=8 \times 8 \mathrm{~cm}^{2}$ $=64 \mathrm{~cm}^{2}$ Now, radius of the quadrant ADQB = 8 cm $\therefore \quad$ Area of the quadrant $\mathrm{ADQB}$ $=\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 8 \times 8 c m^{2}$ $=\frac{1}{4} \times \frac{22}{7} \times 64 \mathrm{~cm}^{2}=\...
Read More →For the reaction
Question: For the reaction $2 \mathrm{~A}_{(g)}+\mathrm{B}_{(g)} \rightarrow 2 \mathrm{D}_{(g)}$ $\Delta U^{\theta}=-10.5 \mathrm{~kJ}$ and $\Delta S^{\ominus}=-44.1 \mathrm{JK}^{-1}$. Calculate $\Delta G^{\theta}$ for the reaction, and predict whether the reaction may occur spontaneously.. Solution: For the given reaction, $2 \mathrm{~A}_{(g)}+\mathrm{B}_{(g)} \rightarrow 2 \mathrm{D}_{(g)}$ $\Delta n_{a}=2-(3)$ = 1 mole Substituting the value of $\Delta U^{\theta}$ in the expression of $\Delta...
Read More →In fig., ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter.
Question: In fig., ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. Solution: $\mathrm{BC}=\sqrt{(14)^{2}+(14)^{2}}=14 \sqrt{2} \mathrm{~cm}$ Area of region II = Area of sector $\mathrm{ABC}$ - Area of $\triangle \mathrm{ABC}=\left\{\frac{1}{4} \pi \times(14)^{2}-\frac{1}{2} \times 14 \times 14\right\} \mathrm{cm}^{2}$ $=\left\{\frac{1}{4} \times \frac{22}{7} \times 196-93\right\} \mathrm{cm}^{2}=56 \mathrm{~cm}^{2}$...
Read More →Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n.
Question: Let $A=\{9,10,11,12,13\}$ and let $t: A \rightarrow N$ be defined by $\{(n)=$ the highest prime factor of $n$. Find the range of $f$. Solution: $A=\{9,10,11,12,13\}$ $f: A \rightarrow \mathbf{N}$ is defined as $f(n)=$ The highest prime factor of $n$ Prime factor of $9=3$ Prime factors of $10=2,5$ Prime factor of $11=11$ Prime factors of $12=2,3$ Prime factor of $13=13$ $\therefore f(9)=$ The highest prime factor of $9=3$ $f(10)=$ The highest prime factor of $10=5$ $f(11)=$ The highest ...
Read More →AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O.
Question: $\mathrm{AB}$ and $\mathrm{CD}$ are respectively arcs of two concentric circles of radii $21 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and centre O. If $\angle \mathrm{AOB}=30^{\circ}$, find the area of the shaded region, Solution: Radius of bigger circle $R=21 \mathrm{~cm}$ and sector angle $\theta=30^{\circ}$ $\therefore$ Area of the sector $\mathrm{OAB}$ $=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^{2}$ $=\frac{11 \times 21}{2} \mathrm{~cm}^{2}=\fra...
Read More →For the reaction,
Question: For the reaction, $2 \mathrm{Cl}_{(g)} \rightarrow \mathrm{Cl}_{2(g)}$, what are the signs of $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ ? Solution: $\Delta H$ and $\Delta S$ are negative The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, $\Delta H$ is negative. Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, $\...
Read More →Show that each of the relation R in the set
Question: Show that each of the relation $\mathrm{R}$ in the $\operatorname{set} \mathrm{A}=\{x \in \mathbf{Z}: 0 \leq x \leq 12\}$, given by (i) $\mathrm{R}=\{(a, b):|a-b|$ is a multiple of 4$\}$ (ii) $\mathrm{R}=\{(a, b): a=b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case. Solution: $A=\{x \in \mathbf{Z}: 0 \leq x \leq 12\}=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$ (i) $\mathrm{R}=\{(a, b):|a-b|$ is a multiple of 4$\}$ For any element $a \in A$, we have $(a, a) \...
Read More →Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}.
Question: Let $f$ be the subset of $Z \times Z$ defined by $f=\{(a b, a+b): a, b \in Z\}$. Is $f$ a function from $Z$ to $Z$ : justify your answer. Solution: The relation $f$ is defined as $f=\{(a b, a+b): a, b \in Z\}$ We know that a relationffrom a set A to a set B is said to be a function if every element of set A has unique images in set B. Since $2,6,-2,-6 \in Z(2 \times 6,2+6),(-2 \times-6,-2+(-6)) \in f$ i.e., $(12,8),(12,-8) \in f$ It can be seen that the same first element i.e., 12 corr...
Read More →In fig., a square OABC is inscribed in a quadrant OPBQ.
Question: In fig., a square $\mathrm{OABC}$ is inscribed in a quadrant $\mathrm{OPBQ}$. If $\mathrm{OA}=20 \mathrm{~cm}$, find the area of the shaded region. (Use $\pi=3.14$ ) Solution: OABC is a square such that its side OA = 20 cm $\therefore \mathrm{OA}=20 \mathrm{~cm}$ $\therefore \mathrm{OB}^{2}=\mathrm{OA}^{2}+\mathrm{AB}^{2}$ $\therefore \mathrm{OB}^{2}=20^{2}+20^{2}$ $=400+400=800$ $\mathrm{OB}=\sqrt{\mathbf{8 0 0}}=\mathbf{2 0} \sqrt{\mathbf{2}} \mathrm{cm}$ Now, area of the quadrant $\...
Read More →Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}.
Question: Let $A=\{1,2,3,4\}, B=\{1,5,9,11,15,16\}$ and $f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}$. Are the following true? (i)fis a relation from A to B (ii)fis a function from A to B. Justify your answer in each case. Solution: $A=\{1,2,3,4\}$ and $B=\{1,5,9,11,15,16\}$ $\therefore A \times B=\{(1,1),(1,5),(1,9),(1,11),(1,15),(1,16),(2,1),(2,5),(2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16),(4,1),(4,5),(4,9),(4,11),(4,15),$, $(4,16)\}$ It is given that $f=\{(1,5),(2,9),(3,1),(4,...
Read More →For the reaction at 298 K,
Question: For the reaction at 298 K, $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$ $\Delta H=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ At what temperature will the reaction become spontaneous considering ΔHand ΔSto be constant over the temperature range? Solution: From the expression, $\Delta G=\Delta H-T \Delta S$ Assuming the reaction at equilibrium, $\Delta T$ for the reaction would be: $T=(\Delta H-\Delta G) \frac{1}{\Delta S...
Read More →In fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm.
Question: In fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region. A Solution: (i) Area of the quadrant OACB $\left(\operatorname{rar}^{-} \mathrm{is}=\frac{\mathbf{7}}{\mathbf{2}} \mathbf{c m}\right)$ $=\frac{1}{4} \times \pi \times r^{2}=\frac{1}{4} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \mathrm{~cm}^{2}$ $=\frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} \mathrm{~cm}^{2}=\frac{11 \time...
Read More →Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}.
Question: Let $R$ be a relation from $N$ to $N$ defined by $R=\left\{(a, b): a, b \in N\right.$ and $\left.a=b^{2}\right\}$. Are the following true $\}$ (i) $(a, a) \in R$, for all $a \in \mathbf{N}$ (ii) $(a, b) \in \mathrm{R}$, implies $(b, a) \in \mathrm{R}$ (iii) $(a, b) \in \mathbf{R},(b, c) \in \mathbf{R}$ implies $(a, c) \in \mathbf{R}$. Justify your answer in each case. Solution: $\mathrm{R}=\left\{(a, b): a, b \in \mathbf{N}\right.$ and $\left.a=b^{2}\right\}$ (i) It can be seen that $2...
Read More →On a square handkerchief, nine circular designs each of radius 7 cm are made.
Question: On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief. Solution: $\because$ The circles touch each other. $\therefore$ The side of the square $A B C D$ = 3 diameter of a circle = 3 (2 radius of a circle) = 3 (2 7 cm) = 42 cm $\rightarrow$ Area of the sollare $A B(\mid)=42 \times 42 \mathrm{~cm}^{2}$ $=1764 \mathrm{~cm}^{2}$ Now, area of one circle $=\pi \mathrm{r}^{2}$ $\Rightarrow \frac{22}{7} \times 7 ...
Read More →Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b,
Question: Let $f=\{(1,1),(2,3),(0,-1),(-1,-3)\}$ be a function from $Z$ to $Z$ defined by $1(x)=a x+b$ for some integers $a, b$. Determine $a, b$. Solution: $f=\{(1,1),(2,3),(0,-1),(-1,-3)\}$ $f(x)=a x+b$ $(1,1) \in f$ $\Rightarrow f(1)=1$ $\Rightarrow a \times 1+b=1$ $\Rightarrow a+b=1$ $(0,-1) \in f$ $\Rightarrow f(0)=-1$ $\Rightarrow a \times 0+b=-1$ $\Rightarrow b=-1$ On substituting $b=-1$ in $a+b=1$, we obtain $a+(-1)=1 \Rightarrow a=1+1=2$. Thus, the respective values of $a$ and $b$ are 2...
Read More →Let f, g: R → R be defined, respectively by f(x) = x + 1,
Question: Let $t, g: \mathbf{R} \rightarrow \mathbf{R}$ be defined, respectively by $1(\mathbf{x})=x+1, g(x)=2 x-3 .$ Find $f+g, f-g$ and $\frac{f}{g}$. Solution: $f, g: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x+1, g(x)=2 x-3$ $(f+g)(x)=f(x)+g(x)=(x+1)+(2 x-3)=3 x-2$ $\therefore(f+g)(x)=3 x-2$ $(f-g)(x)=f(x)-g(x)=(x+1)-(2 x-3)=x+1-2 x+3=-x+4$ $\therefore(f-g)(x)=-x+4$ $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, g(x) \neq 0, x \in \mathbf{R}$ $\therefore\left(\frac{f}{g}\right)...
Read More →Show that the relation R in the set A = {1, 2, 3, 4, 5} given by
Question: Show that the relation R in the setA= {1, 2, 3, 4, 5} given by $\mathrm{R}=\{(a, b):|a-b|$ is even $\}$, is an equivalence relation. Show that all the elements of $\{1,3,5\}$ are related to each other and all the elements of $\{2,4\}$ are related to each other. But no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$. Solution: A= {1, 2, 3, 4, 5} $\mathrm{R}=\{(a, b):|a-b|$ is even $\}$ It is clear that for any element $a \in A$, we have $|a-a|=0$ (which is even). R is refl...
Read More →The area of an equilateral triangle ABC is 17320.5 cm
Question: The area of an equilateral triangle $\mathrm{ABC}$ is $17320.5 \mathrm{~cm}^{2}$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use $\pi=3.14$ and $\sqrt{\mathbf{3}}=1.73205$ ) Solution: Area of the $\triangle \mathrm{ABC}$ (equilateral) $=17320.5 \mathrm{~cm}^{2}$ Let the side of the equilateral $\triangle \mathrm{ABC}$ be $\mathrm{x} \mathrm{cm}$. Then, $\frac{\sqrt{...
Read More →For an isolated system, ΔU = 0, what will be ΔS?
Question: For an isolated system, $\Delta U=0$, what will be $\Delta S ?$ Solution: ΔS will be positive i.e., greater than zero Since $\Delta U=0, \Delta S$ will be positive and the reaction will be spontaneous....
Read More →