Boiling point of water at 750 mm Hg is 99.63°C.
Question: Boiling point of water at 750 mm Hg is 99.63C. How much sucrose is to be added to 500 g of water such that it boils at 100C.Molal elevation constant for water is 0.52 K kg mol1. Solution: Here, elevation of boiling point ΔTb= (100 + 273) (99.63 + 273) = 0.37 K Mass of water,wl= 500 g Molar mass of sucrose (C12H22O11),M2= 11 12 + 22 1 + 11 16 = 342 g mol1 Molal elevation constant,Kb= 0.52 K kg mol1 We know that: $\Delta T_{b}=\frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$ $\...
Read More →In the following figure, shows a sector of a circle,
Question: In the following figure, shows a sector of a circle, centre O, containing an angle0. Prove that: (i) Perimeter of the shaded region is $\left(\tan \theta+\sec \theta+\frac{\pi \theta}{180}-1\right)$ (ii) Area of the shaded region is $\frac{r^{2}}{2}\left(\tan \theta-\frac{\pi \theta}{180}\right)$ Solution: It is given that the radius of circle is $r$ and the angle $\angle A O C=\theta^{\circ}$. In $\triangle A O B$ It is given that $O A=r$. $\cos \theta=\frac{O A}{O B}$ $O B=\frac{O A}...
Read More →Vapour pressure of pure water at 298 K is 23.8 mm Hg.
Question: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Solution: It is given that vapour pressure of water, $p_{1}^{0}=23.8 \mathrm{~mm}$ of $\mathrm{Hg}$ Weight of water taken,w1= 850 g Weight of urea taken,w2= 50 g Molecular weight of water,M1= 18 g mol1 Molecular weight of urea,M2= 60 g mol1 Now, we have to calculate vapour pressure of water in t...
Read More →Prove that
Question: (i) If the vertices of $\Delta A B C$ be $A(1,-3), B(4, p)$ and $C(-9,7)$ and its area is 15 square units, find the values of $p$. (ii) The area of a triangle is $5 \mathrm{sq}$ units. Two of its vertices are $(2,1)$ and $(3,-2)$. If the third vertex is $\left(\frac{7}{2}, y\right)$, find the value of $y$. Solution: (i) Let $A\left(x_{1}, y_{1}\right)=A(1,-3), B\left(x_{2}, y_{2}\right)=B(4, p)$ and $C\left(x_{3}, y_{3}\right)=C(-9,7)$. Now Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\l...
Read More →Solve this
Question: A force $\vec{F}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}$ acts at a point $(4 \hat{i}+3 \hat{j}-\hat{k}) \mathrm{m}$. Then the magnitude of torque about the point $(\hat{i}+2 \hat{j}+\hat{k}) \mathrm{m}$ will be $\sqrt{x} \mathrm{~N}-\mathrm{m}$. The value of $x$ is_____ Solution: $(195)$ Given : $\vec{F}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}$ And, $\vec{r}=[(4 \hat{i}+3 \hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})]=3 \hat{i}+\hat{j}-2 \hat{k}$ Torque, $\tau=\vec{r} \times \vec{F}=...
Read More →The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively,
Question: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. Solution: It is given that: $p_{\mathrm{A}}^{0}=450 \mathrm{~mm}$ of $\mathrm{Hg}$ $p_{\mathrm{B}}^{0}=700 \mathrm{~mm}$ of $\mathrm{Hg}$ ptotal= 600 mm of Hg From Raoults law, we have: $p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$ $p_{\mathrm{B}}=p_{\mathrm{B}}...
Read More →The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively,
Question: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. Solution: It is given that: $p_{\mathrm{A}}^{0}=450 \mathrm{~mm}$ of $\mathrm{Hg}$ $p_{\mathrm{B}}^{0}=700 \mathrm{~mm}$ of $\mathrm{Hg}$ ptotal= 600 mm of Hg From Raoults law, we have: $p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$ $p_{\mathrm{B}}=p_{\mathrm{B}}...
Read More →In a circle of radius 6 cm,
Question: In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110 at the centre of the circle. Find:(i) the circumference of the circle(ii) the area of the circle(iii) the length of the arcAB,(iv) the area of the sectorOAB. Solution: It is given that the radius of circle $r=6 \mathrm{~cm}$, length of chord $=10 \mathrm{~cm}$ and angle at the centre of circle $\theta=110^{\circ}$. (i) We know that the Circumference C of circle of radiusris, $C=2 \pi \mathrm{r}$ $=2 \times \frac{...
Read More →A wheel is rotating freely with an angular speed omega on a shaft.
Question: A wheel is rotating freely with an angular speed $\omega$ on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :$\frac{5}{6}$$\frac{1}{4}$0$\frac{3}{4}$Correct Option: 4, Solution: (4) By angular momentum conservation, $L_{c}=L_{f}$ $\omega I+3 I \times 0=4 I \omega^{\prim...
Read More →The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively,
Question: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. Solution: It is given that: $p_{\mathrm{A}}^{0}=450 \mathrm{~mm}$ of $\mathrm{Hg}$ $p_{\mathrm{B}}^{0}=700 \mathrm{~mm}$ of $\mathrm{Hg}$ ptotal= 600 mm of Hg From Raoults law, we have: $p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$ $p_{\mathrm{B}}=p_{\mathrm{B}}...
Read More →A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD.
Question: $A(6,1), B(8,2)$ and $C(9,4)$ are the vertices of a parallelogram $A B C D$. If $E$ is the midpoint of $D C$, find the area of $\Delta A D E$. Solution: Let $(x, y)$ be the coordinates of $D$ and $\left(x^{\prime}, y^{\prime}\right)$ be the coordinates of $E .$ Since, the diagonals of a parallelogram bisect each other at the same point, therefore $\frac{x+8}{2}=\frac{6+9}{2} \Rightarrow x=7$ $\frac{y+2}{2}=\frac{1+4}{2} \Rightarrow y=3$ Thus, the coordinates ofDare (7, 3).Eis the midpo...
Read More →A sector of 56° cut out from a circle contains
Question: A sector of $56^{\circ}$ cut out from a circle contains area $4.4 \mathrm{~cm}^{2}$. Find the radius of the circle. Solution: We know that the areaAof a sector of circle at an angleof radiusris given by $A=\frac{\theta}{360^{\circ}} \pi r^{2}$ It is given that, Area of a sector $A=4.4 \mathrm{~cm}^{2}$ and angle $\theta=56^{\circ}$. We can find the value ofrby substituting these values in above formula, $A=\frac{56^{\circ}}{360^{\circ}} \times \frac{22}{7} r^{2}$ $4.4=\frac{56^{\circ}}...
Read More →Henry’s law constant for
Question: Henrys law constant for CO2in water is 1.67 108Pa at 298 K. Calculate the quantity of CO2in 500 mL of soda water when packed under 2.5 atm CO2pressure at 298 K. Solution: It is given that: KH= 1.67 108Pa $p_{\mathrm{CO}_{2}}=2.5 \mathrm{~atm}=2.5 \times 1.01325 \times 10^{5} \mathrm{~Pa}$ = 2.533125 105Pa According to Henrys law: $p_{\mathrm{CO}_{2}}=\mathrm{K}_{\mathrm{H}} x$ $\Rightarrow x=\frac{p_{\mathrm{CO}_{2}}}{\mathrm{~K}_{\mathrm{H}}}$ $=\frac{2.533125 \times 10^{5}}{1.67 \tim...
Read More →The minute hand of a clock is 10 cm long.
Question: The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 AM and 8.25 AM. Solution: We know that the areaAof a sector of circle at an angleof radiusris given by $A=\frac{\theta}{360^{\circ}} \pi r^{2}$ We have, Angle described by the minute hand in one minute $=6^{\circ}$ So, Angle described by the minute hand in 25 minute $=6^{\circ} \times 25=150^{\circ}$ $\therefore$ Required area $=\frac{150^{\circ}}{360^{\circ}} \times ...
Read More →For a uniform rectangular sheet shown in the figure,
Question: For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O^{\prime}$ (corner point) is:$2 / 3$$1 / 4$$1 / 8$$1 / 2$Correct Option: , 2 Solution: (2) Moment of inertia of rectangular sheet about an axis passing through $O$, $I_{O}=\frac{M}{12}\left(a^{2}+b^{2}\right)=\frac{M}{12}\left[(80)^{2}+(60)^{2}\right]$ From the parallel axis theorem, moment of inertia about $O^...
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Question: Find the area of $\Delta A B C$ with $A(1,-4)$ and midpoints of sides through $A$ being $(2,-1)$ and $(0,-1)$. Solution: Let $\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ be the coordinates of $B$ and $C$ respectively. Since, the coordinates of $A$ are $(1,-4)$, therefore $\frac{1+x_{2}}{2}=2 \Rightarrow x_{2}=3$ $\frac{-4+y_{2}}{2}=-1 \Rightarrow y_{2}=2$ $\frac{1+x_{3}}{2}=0 \Rightarrow x_{3}=-1$ $\frac{-4+y_{3}}{2}=-1 \Rightarrow y_{3}=2$ Let $A\left(x_{1}, y_{1}\right...
Read More →The minute hand of a clock is
Question: The minute hand of a clock is $\sqrt{21} \mathrm{~cm}$ long. Find the area described by the minute hand on the face of the clock between $7.00 \mathrm{AM}$ and $7.05 \mathrm{AM}$. Solution: We know that the areaAof a sector of circle at an angleof radiusris given by $A=\frac{\theta}{360^{\circ}} \pi r^{2}$ We have, Angle described by the minute hand in one minute $=6 \%$ So, angle described by the minute hand in five minute $=6^{\circ} \times 5=30^{\circ}$ Thus, Area swept by the minut...
Read More →Solve the following
Question: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henrys law constant. Solution: It is given thatthe solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water. Moles of water $=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}$ = 55.56 mol $\therefore$ Mole fraction of $\mathrm{H}_{2} \mathrm{~S}, x=\frac{\text { Moles of } \mathrm{H}_{2} \...
Read More →A(7, −3), B(5, 3), C(3, −1) are the vertices of
Question: $A(7,-3), B(5,3), C(3,-1)$ are the vertices of a $\Delta A B C$ and $A D$ is its median. Prove that the median $A D$ divides $\Delta A B C$ into two triangles of equal areas. Solution: The vertices of the triangle areA(7, 3),B(5, 3),C(3, 1). Coordinates of $D=\left(\frac{5+3}{2}, \frac{3-1}{2}\right)=(4,1)$ For the area of the triangle $A D C$, let $A\left(x_{1}, y_{1}\right)=A(7,-3), D\left(x_{2}, y_{2}\right)=D(4,1)$ and $C\left(x_{3}, y_{3}\right)=C(3,-1)$. Then Area of $\Delta A D ...
Read More →In a circle of radius 21 cm,
Question: In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. Find (i) the length of the arc (ii) area of the sector formed by the arc. (Use = 22/7) Solution: Here, we have = 60 andr= 21 cm(i) The length of the arc is given by $\frac{60^{\circ}}{360^{\circ}} \times 2 \pi(21)$ $=\frac{1}{6} \times 2 \times \frac{22}{7} \times 21$ $=22 \mathrm{~cm}$ (ii) Area of the sector formed by the arc is given by $\frac{60^{\circ}}{360^{-}} \pi(21)^{2}$ $=\frac{1}{6} \times \frac{22}{7...
Read More →Calculate (a) molality (b) molarity and (c) mole fraction of
Question: Calculate(a)molality(b)molarity and(c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1. Solution: (a)Molar mass of KI = 39 + 127 = 166 g mol1 20% (mass/mass) aqueoussolution of KI means 20 g of KI is present in 100 g of solution. That is, 20 g of KI is present in (100 20)g of water = 80 g of water Therefore, molality of the solution $=\frac{\text { Moles of KI }}{\text { Mass of water in } \mathrm{kg}}$ $=\frac{\frac{20}{166}}{0.08} \mathrm{~m}$ = 1.506 ...
Read More →Consider two uniform discs of the same thickness
Question: Consider two uniform discs of the same thickness and different radii $R_{1}=R$ and $R_{2}=\alpha R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$, respectively, about their axes is $I_{1}: I_{2}=1: 16$ then the value of $\alpha$ is:$2 \sqrt{2}$$\sqrt{2}$24Correct Option: , 3 Solution: (3) Let $p$ be the density of the discs and $t$ is the thickness of discs. Moment of inertia of disc is given by $I=\frac{M R^{2}}{2}=\frac{\left[\rho\left(\pi R^...
Read More →The length of the minute hand of a clock is 14 cm.
Question: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. Solution: Angle make by the minute hand in 1 minute = 6∘ Angle make by the minute hand in 5 minute = 5 ⨯ 6∘= 30∘ Area of the sector having central angle is given by $\frac{30^{\circ}}{360^{\circ}} \pi(14)^{2}$ $=\frac{1}{12} \times \frac{22}{7}(14)^{2}$ $=51.33 \mathrm{~cm}^{2}$ Hence, the area swept by minute hand in 5 minutes is 51.33 cm2...
Read More →A sector is cut-off from a circle of radius 21 cm.
Question: A sector is cut-off from a circle of radius 21 cm. The angle of the sector is 120. Find the length of its arc and the area. Solution: We know that the arc length / and area $A$ of a sector of circle at an angle $\theta$ of radius $r$ is given by $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ and angle $A=\frac{\theta}{360^{\circ}} \pi r^{2}$. Let OAB is the given sector. It is given that $O A=21 \mathrm{~cm}$ and angle $\angle A O B=120^{\circ}$. Now using the value ofrand,we will find ...
Read More →Find area of the triangle formed by joining the midpoints of the sides of the triangle
Question: Find area of the triangle formed by joining the midpoints of the sides of the trianglewhose vertices areA(2, 1),B(4, 3) andC(2, 5). Solution: The vertices of the triangle areA(2, 1),B(4, 3) andC(2, 5). Coordinates of midpoint of $A B=P\left(x_{1}, y_{1}\right)=\left(\frac{2+4}{2}, \frac{1+3}{2}\right)=(3,2)$ Coordinates of midpoint of $B C=Q\left(x_{2}, y_{2}\right)=\left(\frac{4+2}{2}, \frac{3+5}{2}\right)=(3,4)$ Coordinates of midpoint of $A C=R\left(x_{3}, y_{3}\right)=\left(\frac{2...
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