A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius.
Question: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$. Solution: Here, r = 1 cm and h = 1 cm. Volume of the conical part $=\frac{\mathbf{1}}{\mathbf{3}} \pi r^{2} h$ and volume of the hemispherical part $\frac{2}{3} \pi \mathrm{r}^{3}$ $\therefore$ Volume of the solid shape $=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}=\frac{1}{3} \pi r^{2}[...
Read More →For the following equilibrium,
Question: For the following equilibrium, $K_{\mathrm{C}}=6.3 \times 10^{14}$ at $1000 \mathrm{~K}$ $\mathrm{NO}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \longleftrightarrow \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What isKc, for the reverse reaction? Solution: It is given that $K_{C}$ for the forward reaction is $6.3 \times 10^{14}$. Then, $K_{C}$ for the reverse reaction will be, $K_{...
Read More →Find the degree measure of the angle subtended at the
Question: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm$\left(\right.$ Use $\left.\pi=\frac{22}{7}\right)$ Solution: We know that in a circle of radiusrunit, if an arc of lengthlunit subtends an angleradian at the centre, then $\theta=\frac{1}{r}$ Therefore, forr $=100 \mathrm{~cm}, \mathrm{l}=22 \mathrm{~cm}$, we have $\theta=\frac{22}{100}$ radian $=\frac{180}{\pi} \times \frac{22}{100}$ deg ree $=\frac{180 \times 7 \times 2...
Read More →Let R be the relation in the set {1, 2, 3, 4} given by R
Question: Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer. A) R is reflexive and symmetric but not transitive. (B) R is reflexive and transitive but not symmetric. (C) R is symmetric and transitive but not reflexive. (D) R is an equivalence relation. Solution: $R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ It is seen that $(a, a) \in \mathbf{R}$, for every $a \in\{1,2,3,4\}$. R is reflexive. It is s...
Read More →A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in fig.
Question: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in fig. If the height of the cylinder is 10 cm, and its base is of radius 35 cm, find the total surface area of the article. Solution: Radius of the cylinder (r) = 3.5 cm Height of the cylinder (h) = 10 cm $\therefore$ Curved surface area of cylinder $=2 \pi \mathrm{rh}$ $=2 \times \frac{22}{7} \times \frac{35}{10} \times 10 \mathrm{~cm}^{2}=220 \mathrm{~cm}^{2}$ Curved surface area of a ...
Read More →A wheel makes 360 revolutions in one minute.
Question: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Solution: Number of revolutions made by the wheel in 1 minute = 360 \thereforeNumber of revolutions made by the wheel in 1 second $=\frac{360}{60}=6$ In one complete revolution, the wheel turns an angle of $2 \pi$ radian. Hence, in 6 complete revolutions, it will turn an angle of $6 \times 2 \pi$ radian, i.e., $12 \pi$ radian Thus, in one second, the wheel turns an angle of $12 \pi$ radian...
Read More →From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm,
Question: From a solid cylinder whose height is $2.4 \mathrm{~cm}$ and diameter $1.4 \mathrm{~cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\mathrm{cm}^{2}$. Solution: For cylinder part : Height = 2.4 cm and diameter = 1.4 cm $\Rightarrow$ Radius $(r)=0.7 \mathrm{~cm}$ $\therefore$ Total surface area of the cylindrical part $=2 \times \frac{\mathbf{2 2}}{\mathbf{7}} \times \frac{\mathbf{7}}{\mathbf{...
Read More →Find out the value of Kc for each of the following equilibria from the value of Kp:
Question: Find out the value ofKcfor each of the following equilibria from the value ofKp: (i) $\quad 2 \mathrm{NOCl}(\mathrm{g}) \longleftrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \quad K_{p}=1.8 \times 10^{-2}$ at $500 \mathrm{~K}$ (ii) $\mathrm{CaCO}_{3}$ (s) $\longleftrightarrow \mathrm{CaO}$ (s) $+\mathrm{CO}_{2}$ (g); $\quad \mathrm{K}_{p}=167$ at $1073 \mathrm{~K}$ Solution: The relation between $K_{p}$ and $K_{c}$ is given as: $K_{p}=K_{c}(\mathrm{RT})^{\Delta n...
Read More →Let L be the set of all lines in XY plane and R be the relation in L defined as R
Question: Let $L$ be the set of all lines in $X Y$ plane and $R$ be the relation in $L$ defined as $R=\left\{\left(L_{1}, L_{2}\right)\right.$ : $L_{1}$ is parallel to $\left.L_{2}\right\}$. Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y=2 x+4$. Solution: $R=\left\{\left(L_{1}, L_{2}\right): L_{1}\right.$ is parallel to $\left.L_{2}\right\}$ $R$ is reflexive as any line $L_{1}$ is parallel to itself i.e., $\left(L_{1}, L_{1}\right) \in R$. Now, Let $\l...
Read More →Write the expression for the equilibrium constant, Kc for each of the following
Question: Write the expression for the equilibrium constant,Kcfor each of the following reactions: (i) $2 \mathrm{NOCl}(\mathrm{g}) \longleftrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})$ (ii) $2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \longleftrightarrow 2 \mathrm{CuO}(\mathrm{s})+4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ (iii) $\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longleftrigh...
Read More →A tent is in the shape of a cylinder surmounted by a conical top.
Question: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively, and the slant height of the top is $2.8 \mathrm{~m}$, find the area of the canvas used for making the tent. Also find the cost of the canvas of the tent at the rate of Rs. 500 per $\mathrm{m}^{2}$ (Note that the base of the tent will not be covered with canvas). Solution: Radius of the cylindrical base = 2m and heig...
Read More →Find the degree measures corresponding to the following radian measures
Question: Find the degree measures corresponding to the following radian measures $\left(\right.$ Use $\left.\pi=\frac{22}{7}\right)$ (i) $\frac{11}{16}$ (ii) $-4$ (iii) $\frac{5 \pi}{3}$ (iv) $\frac{7 \pi}{6}$ Solution: (i) $\frac{11}{16}$ We know that $\pi$ radian $=180^{\circ}$ $\therefore \frac{11}{16}$ radain $=\frac{180}{\pi} \times \frac{11}{16}$ deg ree $=\frac{45 \times 11}{\pi \times 4}$ deg ree $=\frac{45 \times 11 \times 7}{22 \times 4}$ deg ree $=\frac{315}{8}$ deg ree $=39 \frac{3}...
Read More →Show that the relation R defined in the set A of all polygons as R
Question: Show that the relation $R$ defined in the set $A$ of all polygons as $R=\left\{\left(P_{1}, P_{2}\right): P_{1}\right.$ and $P_{2}$ have same number of sides , is an equivalence relation. What is the set of all elements in $A$ related to the right angle triangle $T$ with sides 3,4 and 5 ? Solution: $\mathrm{R}=\left\{\left(P_{1}, P_{2}\right): P_{1}\right.$ and $P_{2}$ have same the number of sides $\}$ $\mathrm{R}$ is reflexive since $\left(P_{1}, P_{1}\right) \in \mathrm{R}$ as the s...
Read More →At a certain temperature and total pressure of 105
Question: At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms $\mathrm{I}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{I}(\mathrm{g})$ CalculateKpfor the equilibrium. Solution: Partial pressure of I atoms, $p_{1}=\frac{40}{100} \times p_{\text {total }}$ $=\frac{40}{100} \times 10^{5}$ $=4 \times 10^{4} \mathrm{~Pa}$ Partial pressure of I2molecules, $p_{\mathrm{l}_{2}}=\frac{60}{100} \times p_{\text {total }}$ $=\frac{60}{100} \times 10^{5}$ $=...
Read More →A medicine capsule is in the shape of a cylinder with two hemispheres struck to each of its ends (see figure).
Question: A medicine capsule is in the shape of a cylinder with two hemispheres struck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. Solution: Surface area of the cylindrical part $=2 \pi \times \mathrm{r} \times \mathrm{h}$ $=2 \pi \times\left(\frac{\mathbf{5}}{\mathbf{2}}\right) \times 9 \mathrm{~mm}^{2}=45 \pi \mathrm{mm}^{2}$ Sum of the curved surface areas of two hemispherical parts. $=2\left\{\mat...
Read More →A hemispherical depression is cut out from one face of a cubical wooden block such
Question: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $\ell$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. Solution: Let $\ell$ be the side of the cube. $\therefore$ The greatest diameter of the hemisphere $=\ell$ $\Rightarrow$ Radius of the hemisphere $=\frac{\ell}{\mathbf{2}}$ $\therefore$ Surface area of hemisphere $=2 \pi \mathrm{r}^{2}$ $=2 \times \pi \times \frac{\ell}{\mathbf{...
Read More →What is Kc for the following equilibrium when the equilibrium concentration of each substance is
Question: What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $\left[\mathrm{SO}_{2}\right]=0.60 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.82 \mathrm{M}$ and $\left[\mathrm{SO}_{3}\right]=1.90 \mathrm{M} ?$ $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})$ Solution: The equilibrium constant (Kc) for the give reaction is: Hence, $K_{c}$ for the equilibrium is $12.239 \mathrm{M}^{-1}$....
Read More →Show that the relation R defined in the set A of all triangles as
Question: Show that the relation R defined in the setAof all triangles as Show that the relation $R$ defined in the set $A$ of all triangles as $R=\left\{\left(T_{1}, T_{2}\right): T_{1}\right.$ is similar to $\left.T_{2}\right\}$, is equivalence relation. Consider three right angle triangles $T_{1}$ with sides $3,4,5, T_{2}$ with sides $5,12,13$ and $T_{3}$ with sides $6,8,10$. Which triangles among $T_{1}, T_{2}$ and $T_{3}$ are related? Solution: $R=\left\{\left(T_{1}, T_{2}\right): T_{1}\rig...
Read More →Find the degree measures corresponding to the following radian measures
Question: Find the degree measures corresponding to the following radian measures $\left(\right.$ Use $\left.\pi=\frac{22}{7}\right)$ (i) $\frac{11}{16}$ (ii) $-4$ (iii) $\frac{5 \pi}{3}$ (iv) $\frac{7 \pi}{6}$ Solution: (i) $\frac{11}{16}$ We know that $\pi$ radian $=180^{\circ}$ $\therefore \frac{11}{16}$ radain $=\frac{180}{\pi} \times \frac{11}{16}$ deg ree $=\frac{45 \times 11}{\pi \times 4}$ deg ree $=\frac{45 \times 11 \times 7}{22 \times 4}$ deg ree $=\frac{315}{8}$ deg ree $=39 \frac{3}...
Read More →A cubical block of side 7 cm is surmounted by a hemisphere.
Question: A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. Solution: On 7 cm 7 cm base of the cubical block, we can mount hemisphere having greatest diameter equal to 7 cm. Here, the radius of the hemisphere = 3.5 cm. Now, the surface area of the solid made in figure. = The surface area of the cube + The curved surface area of the hemisphere The area of the base of the hemisphere. $=\left\{6 \ti...
Read More →A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
Question: A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure? Solution: (a)If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This ...
Read More →A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.
Question: A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure? Solution: (a)If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This ...
Read More →A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.
Question: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. find the total surface area of the toy Solution: Let r and h be the radius of cone, hemisphere and height of cone $\therefore \mathrm{h}=(15.5-3.5) \mathrm{cm}$ $=12.0 \mathrm{~cm}$ Also $\ell^{2}=\mathrm{h}^{2}+\mathrm{r}^{2}$ $=12^{2}+(3.5)^{2}$ $=156.25$ $\therefore \ell=12.5 \mathrm{~cm}$ Curved surface area of the conical part $=\pi r \ell$ Curved surface...
Read More →Show that the relation R in the set A of points in a plane given by
Question: Show that the relation $R$ in the set $A$ of points in a plane given by $R=\{(P, Q)$ : distance of the point $P$ from the origin is same as the distance of the point $Q$ from the origin $\}$, is an equivalence relation. Further, show that the set of all point related to a point $P \neq(0,0)$ is the circle passing through $P$ with origin as centre. Solution: $R=\{(P, Q)$ : distance of point $P$ from the origin is the same as the distance of point $Q$ from the origin $\}$ Clearly, $(P, P...
Read More →Find the radian measures corresponding to the following degree measures:
Question: Find the radian measures corresponding to the following degree measures: (i) 25 (ii) 47 30' (iii) 240 (iv) 520 Solution: (i) $25^{\circ}$ We know that $180^{\circ}=\pi$ radian $\therefore 25^{\circ}=\frac{\pi}{180} \times 25$ radian $=\frac{5 \pi}{36}$ radian (ii) $-47^{\circ} 30^{\prime}$ $-47^{\circ} 30^{\prime}=-47 \frac{1}{2}$ degree $\left[1^{\circ}=60^{\prime}\right]$ $=\frac{-95}{2}$ degree Since $180^{\circ}=\pi$ radian $\frac{-95}{2}$ deg ree $=\frac{\pi}{180} \times\left(\fra...
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