Calculate the mass of urea
Question: Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. Solution: Molar mass of urea (NH2CONH2) = 2(1 14 + 2 1) + 1 12 + 1 16 = 60 g mol1 0.25 molar aqueous solution of urea means: 1000 g of water contains 0.25 mol = (0.25 60)g of urea = 15 g of urea That is, (1000 + 15) g of solution contains 15 g of urea Therefore, $2.5 \mathrm{~kg}(2500 \mathrm{~g})$ of solution contains $=\frac{15 \times 2500}{1000+15} \mathrm{~g}$ = 36.95 g = 37 g of urea (a...
Read More →A circular disc of mass M and radius R is rotating about
Question: A circular disc of mass $M$ and radius $R$ is rotating about its axis with angular speed $\omega_{1}$. If another stationary disc having radius $\frac{R}{2}$ and same mass $M$ is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed $\omega_{2}$. The energy lost in the process is $p \%$ of the initial energy. Value of $p$ is__________ Solution: (20) As we know moment of inertia disc, $I_{\mathrm{disc}}=\frac{1}{2} M R^{2}$ Using angular momentum...
Read More →The perimeter of a certain sector of a circle of radius
Question: The perimeter of a certain sector of a circle of radius 5.6 m is 27.2 m. Find the area of the sector. Solution: We know that the areaAof a sector of circle of radiusrand arc lengthlis given by $A=\frac{1}{2} l r$ Let OAB is the given sector. Then, Perimeter of sector OAB = 27.2 m OA + OB + arc AB = 27.2 m 5.6 + 5.6 + arc AB = 27.2 m 11.2 + arc AB = 27.2 m $\operatorname{arc} \mathrm{AB}=16 \mathrm{~m}$ So, $l=16 \mathrm{~m}$ Now substituting the value ofrandl in above formula, $A=\frac...
Read More →If A(–7, 5), B(–6, –7), C(–3, –8) and D(2, 3) are the vertices of a quadrilateral
Question: IfA(7, 5),B(6, 7),C(3, 8) andD(2, 3) are the vertices of a quadrilateralABCDthen find the area of the quadrilateral. Solution: Consider the figure. Construction: Produce $A C$ by joining points $A$ to $C$ to form two triangles, $\Delta A B C$ and $\Delta A D C$. In $\Delta \mathrm{ABC}$, $x_{1}=-7, x_{2}=-6$ and $x_{3}=-3 ; y_{1}=5, y_{2}=-7$ and $y_{3}=-8$ We know that, $\operatorname{ar}(\Delta \mathrm{ABC})=\left|\frac{1}{2}\left\{\left(x_{1}\right)\left(y_{2}-y_{3}\right)+\left(x_{...
Read More →Calculate the molarity of each of the following solutions:
Question: Calculate the molarity of each of the following solutions:(a)30 g of Co(NO3)2. 6H2O in 4.3 L of solution(b)30 mL of 0.5 M H2SO4diluted to 500 mL. Solution: Molarity is given by: Molarity $=\frac{\text { Moles of solute }}{\text { Volume of solution in litre }}$ (a)Molar mass of Co (NO3)2.6H2O = 59 + 2 (14 + 3 16) + 6 18 = 291 g mol1 $\therefore$ Moles of $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}=\frac{30}{291} \mathrm{~mol}$ = 0.103 mol Therefore, m...
Read More →The perimeter of a sector of a circle of radius
Question: The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector. Solution: We know that the area A of a sector of circle of radius r and arc length l is given by $A=\frac{1}{2} l r$ Let OAB is the given sector. Then, Perimeter of sector $O A B=27.2$ $O A+O B+\operatorname{arc} A B=27.2$ $5.7+5.7+\operatorname{arc} A B=27.2$ $11.4+\operatorname{arc} A B=27.21$ $\operatorname{arc} A B=15.8 \mathrm{~m}$ So, $I=15.8 \mathrm{~m}$ Now substituting the value ofran...
Read More →Calculate the mole fraction of benzene in solution containing 30%
Question: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Solution: Let the total mass of the solution be 100 g and the mass of benzene be 30 g. Mass of carbon tetrachloride = (100 30)g = 70 g Molar mass of benzene (C6H6) = (6 12 + 6 1) g mol1 = 78 g mol1 $\therefore$ Number of moles of $\mathrm{C}_{6} \mathrm{H}_{6}=\frac{30}{78} \mathrm{~mol}$ = 0.3846 mol Molar mass of carbon tetrachloride (CCl4) = 1 12 + 4 35.5 = 154 g mol1 $\therefore$ Numb...
Read More →In a circle of radius 35 cm,
Question: In a circle of radius 35 cm, an arc subtends an angle of 72 at the centre. Find the length of the arc and area of the sector. Solution: We know that the arc length / and area $A$ of a sector of an angle $\theta$ in the circle of radius $r$ is given by $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ and $A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$ respectively. It is given that, $r=35 \mathrm{~cm}$ and $\theta=72^{\circ}$. We will calculate the arc length using the value ofrand, $l=\fr...
Read More →Calculate the mass percentage of benzene
Question: Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Solution: Mass percentage of $\mathrm{C}_{6} \mathrm{H}_{6}=\frac{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}}{\text { Total mass of the solution }} \times 100 \%$ $=\frac{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}}{\text { Mass of } \mathrm{C}_{6} \mathrm{H}_{6}+\text { Mass of } \mathrm{CCl}_{4}} \times 100 \%$ $=\frac{22}{22+122} ...
Read More →Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).
Question: Find the area of quadrilateralABCDwhose vertices areA(3, 1),B(2, 4),C(4, 1) andD(3, 4). Solution: By joiningAandC, we get two trianglesABCandACD. Let $A\left(x_{1}, y_{1}\right)=A(-3,-1), B\left(x_{2}, y_{2}\right)=B(-2,-4), C\left(x_{3}, y_{3}\right)=C(4,-1)$ and $D\left(x_{4}, y_{4}\right)=D(3,4)$. Then Area of $\Delta A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[-3(-4+1)-2(-1+1)+4(-1+4)]$ $=\fr...
Read More →A B C is a plane lamina of the shape of an equilateral triangle.
Question: $A B C$ is a plane lamina of the shape of an equilateral triangle. $D, E$ are mid points of $A B, A C$ and $G$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $G$ and perpendicular to the plane $A B C$ is $I_{0}$. If part $A D E$ is removed, the moment of inertia of the remaining part about the same axis is $\frac{N I_{0}}{16}$ where $N$ is an integer. Value of $N$ is______ Solution: (11) Let mass of triangular lamina $=m$, and length of sid...
Read More →AB is a chord of a circle with centre O and radius 4 cm.
Question: ABis a chord of a circle with centreOand radius 4 cm.ABis of length 4 cm. Find the area of the sector of the circle formed by chordAB. Solution: We have to find the area of the sector AOB formed by the chord AB. We have $O A=4 \mathrm{~cm}$ and $A B=4 \mathrm{~cm}$. So, $A L=\frac{A B}{2} \mathrm{~cm}$ $=\frac{4}{2} \mathrm{~cm}$ $=2 \mathrm{~cm}$ Let $\angle A O B=2 \theta$. Then, $\angle A O L=\angle B O L$ $=\theta$ In $\triangle O L A$, we have $\sin \theta=\frac{A L}{O A}$ =\frac{...
Read More →Explain the following with suitable examples:
Question: Explain the following with suitable examples: (i)Ferromagnetism (ii)Paramagnetism (iii)Ferrimagnetism (iv)Antiferromagnetism (v)12-16 and 13-15 group compounds. Solution: (i)Ferromagnetism:The substances that are strongly attracted by a magnetic field are called ferromagnetic substances. Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and CrO2. In solid state,...
Read More →Explain the following with suitable examples:
Question: Explain the following with suitable examples: (i)Ferromagnetism (ii)Paramagnetism (iii)Ferrimagnetism (iv)Antiferromagnetism (v)12-16 and 13-15 group compounds. Solution: (i)Ferromagnetism:The substances that are strongly attracted by a magnetic field are called ferromagnetic substances. Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and CrO2. In solid state,...
Read More →The area of a sector of a circle of radius 5 cm
Question: The area of a sector of a circle of radius $5 \mathrm{~cm}$ is $5 \pi \mathrm{cm}^{2}$. Find the angle contained by the sector. Solution: We know that the areaAof a sector of an anglein the circle of radiusris given by $A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$ It is given that radius $r=5 \mathrm{~cm}$ and area $A=5 \pi \mathrm{cm}^{2}$. Now we substitute the value ofrandAin above formula to find the value of, $5 \pi=\frac{\theta}{360^{\circ}} \times \pi \times 5 \times 5$ $\thet...
Read More →The area of a sector of a circle of radius 2 cm
Question: The area of a sector of a circle of radius $2 \mathrm{~cm}$ is $\pi \mathrm{cm}^{2}$. Find the angle contained by the sector. Solution: We know that the areaAof a sector of an anglein the circle of radiusris given by $A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$ It is given that $r=2 \mathrm{~cm}$ and area $A=\pi \mathrm{cm}^{2}$. Now we substitute the value ofrandAin above formula to find the value of, $\pi=\frac{\theta}{360^{\circ}} \times \pi \times 2 \times 2$ $\theta=\frac{360^{...
Read More →An massless equilateral triangle E F G$ of side ' a ' (As shown in figure)
Question: An massless equilateral triangle $E F G$ of side ' $a$ ' (As shown in figure) has three particles of mass $m$ situated at its vertices. The moment of inertia of the system about the line $E X$ perpendicular to $E G$ in the plane of $E F G$ is $\frac{N}{20} m a^{2}$ where $N$ is an integer. The value of $N$ is Solution: (25) Moment of inertia of the system about axis $X E$. $I=I_{E}+I_{F}+I_{G}$ $\Rightarrow I=m\left(r_{E}\right)^{2}+m\left(r_{F}\right)^{2}+m\left(r_{G}\right)^{2}$ $\Ri...
Read More →A sector of a circle of radius 8 cm contains an angle of 135°.
Question: A sector of a circle of radius 8 cm contains an angle of 135. Find the area of the sector. Solution: We know that the areaAof a sector of an anglein the circle of radiusris given by $A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$ It is given that $r=8 \mathrm{~cm}$ and $\theta=135^{\circ}$. Now we substitute the value of $r$ and $\theta$ in above formula, $A=\frac{135^{\circ}}{360^{\circ}} \times \pi \times 8 \times 8 \mathrm{~cm}^{2}$ $=24 \pi \mathrm{cm}^{2}$...
Read More →If NaCl is doped with
Question: If NaCl is doped with 103mol % of SrCl2, what is the concentration of cation vacancies? Solution: It is given thatNaCl is doped with 103mol% of SrCl2. This means that100 mol of NaCl is doped with 103mol of SrCl2. Therefore, $1 \mathrm{~mol}$ of $\mathrm{NaCl}$ is doped with $\frac{10^{-3}}{100} \mathrm{~mol}$ of $\mathrm{SrCl}_{2}$ = 105mol of SrCl2 Cation vacancies produced by one Sr2+ion = 1 $\therefore$ Concentration of the cation vacancies produced by $10^{-5} \mathrm{~mol}$ of $\m...
Read More →Find the area of quadrilateral PQRS whose vertices are
Question: Find the area of quadrilateralPQRSwhose vertices areP(5, 3),Q(4, 6),R(2, 3) andS(1, 2). Solution: By joiningPandR, we get two trianglesPQRandPRS. Let $P\left(x_{1}, y_{1}\right)=P(-5,-3), Q\left(x_{2}, y_{2}\right)=Q(-4,-6), R\left(x_{3}, y_{3}\right)=R(2,-3)$ and $S\left(x_{4}, y_{4}\right)=S(1,2)$. Then Area of $\Delta P Q R=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[-5(-6+3)-4(-3+3)+2(-3+6)]$ $=\fr...
Read More →A sector of a circle of radius 4 cm contains an angle of 30°.
Question: A sector of a circle of radius 4 cm contains an angle of 30. Find the area of the sector. Solution: We know that the areaAof a sector of an anglein the circle of radiusris given by $A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$ It is given that $r=4 \mathrm{~cm}$ and angle $\theta=30^{\circ}$. Now we substitute the value ofrandin above formula, $A=\frac{30^{\circ}}{360^{\circ}} \times \pi \times 4 \times 4 \mathrm{~cm}^{2}$ $=\frac{4 \pi}{3} \mathrm{~cm}^{2}$...
Read More →Aluminium crystallises in a cubic close-packed structure.
Question: Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (i)What is the length of the side of the unit cell? (ii)How many unit cells are there in 1.00 cm3of aluminium? Solution: (i)For cubic close-packed structure: $a=2 \sqrt{2} r$ $=2 \sqrt{2} \times 125 \mathrm{pm}$ = 353.55 pm = 354 pm (approximately) (ii)Volume of one unit cell = (354 pm)3 = 4.4 107pm3 = 4.4 107 1030cm3 = 4.4 1023cm3 Therefore, number of unit cells in $1.00 \mathrm{~cm}^{3}=\frac{1.0...
Read More →Find the angle subtended at the centre of a
Question: Find the angle subtended at the centre of a circle of radius 'a' by an arc of length (a/4) cm. Solution: We know that the arc lengthlof a sector of an anglein a circle of radiusris $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ It is given $l=\frac{a \pi}{4} \mathrm{~cm}$ and radius $r=a \mathrm{~cm}$. Now we substitute the value oflandrin above formula to find the value of anglesubtended at the centre of circle. $\frac{a \pi}{4} \mathrm{~cm}=\frac{\theta}{360^{\circ}} \times 2 \pi \tim...
Read More →An arc of length 15 cm subtends an angle of 45°
Question: An arc of length 15 cm subtends an angle of 45 at the centre of a circle. Find in terms of , the radius of the circle. Solution: We know that the arc lengthlof a sector of an anglein a circle of radius r is $l=\frac{\theta}{360^{\circ}} \times 2 \pi r$ It is given that $l=15 \mathrm{~cm}$ and angle $\theta=45^{\circ}$. Now we substitute the value oflandin above formula to find the value of radiusrof circle. $15 \mathrm{~cm}=\frac{45^{\circ}}{360^{\circ}} \times 2 \pi r$ $r=\frac{15 \ti...
Read More →Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5),
Question: Find the area of quadrilateralABCDwhose vertices areA(3, 1),B(9, 5),C(14, 0) andD(9, 19). Solution: By joiningAandC, we get two trianglesABCandACD. Let $A\left(x_{1}, y_{1}\right)=A(3,-1), B\left(x_{2}, y_{2}\right)=B(9,-5), C\left(x_{3}, y_{3}\right)=C(14,0)$ and $D\left(x_{4}, y_{4}\right)=D(9,19)$. Then Area of $\Delta A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[3(-5-0)+9(0+1)+14(-1+5)]$ $=\fr...
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