In the given figure, side BC of ∆ABC is bisected at D and O is any point on AD.
Question: In the given figure, sideBCof ∆ABCis bisected atDandOis any point onAD.BOandCOproduced meetACandABatEandF, respectively, andADis produced toX, so thatDis the midpoint ofOX. Prove thatAO:AX=AF:ABand show thatEF∥BC. Solution: Join BX and CX.It is given that BC is bisected at D.BD=DCIt is also given thatOD=OXThe diagonalsOXandBCof quadrilateralBOCX bisect each other.Therefore,BOCXis a parallelogram. $\therefore B O \| C X$ and $B X \| C O$ $B X \| C F$ and $C X \| B E$ $B X \| O F$ and $...
Read More →If a wire is bent into the shape of a square,
Question: If a wire is bent into the shape of a square, then the area of the square is 81 cm2. When wire is bent into a semi-circular shape, then the area of the semi-circle will be (a) $22 \mathrm{~cm}^{2}$ (b) $44 \mathrm{~cm}^{2}$ (c) $77 \mathrm{~cm}^{2}$ (d) $154 \mathrm{~cm}^{2}$ Solution: We have given that a wire is bent in the form of square of side a cm such that the area of the square is. If we bent the same wire in the form of a semicircle with radiusrcm, the perimeter of the wire wi...
Read More →The rate constant for the decomposition of hydrocarbons is
Question: The rate constant for the decomposition of hydrocarbons is 2.418 105s1at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor. Solution: k= 2.418 105s1 T= 546 K Ea= 179.9 kJ mol1= 179.9 103J mol1 According to the Arrhenius equation, $k=\mathrm{Ae}^{-E_{a} / \mathrm{R} T}$ $\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$ $\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{R} T}$ $\Rightarrow \log \mathrm{A}=\log k+\...
Read More →The rate constant for the decomposition of
Question: The rate constant for the decomposition of N2O5at various temperatures is given below: Draw a graph between lnkand 1/Tand calculate the values ofAandEa. Predict the rate constant at 30 and 50C. Solution: From the given data, we obtain Slope of the line, According to Arrhenius equation, Slope $=-\frac{E_{a}}{R}$ $\Rightarrow E_{a}=-S$ lope $\times \mathrm{R}$ $=-(-12.301 \mathrm{~K}) \times\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ $=102.27 \mathrm{~kJ} \mathrm{...
Read More →A wire can be bent in the form of a circle of radius 56 cm.
Question: A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form fo a square, then its area will be (a) $3520 \mathrm{~cm}^{2}$ (b) $6400 \mathrm{~cm}^{2}$ (c) $7744 \mathrm{~cm}^{2}$ (d) $8800 \mathrm{~cm}^{2}$ Solution: We have given that a wire is bent in the form of circle of radius 56 cm. If we bent the same wire in the form of square of side a cm, the perimeter of the wire will not change. $\therefore$ perimeter of the circle $=$ perimeter of the square $\the...
Read More →∆ABC and ∆DBC lie on the same side of BC, as shown in the figure.
Question: ∆ABCand ∆DBClie on the same side ofBC, as shown in the figure. From a pointPonBC,PQ∥ABandPR∥BDare drawn, meetingACatQandCDatR, respectively. Prove thatQR∥AD. Solution: In $\triangle C A B, P Q \| A B$. Applying Thales' theorem, we get: $\frac{C P}{P B}=\frac{C Q}{Q A} \quad \ldots(1)$ Similarly, applying Thales' theorem in $\triangle B D C$, where $P R \| B D$, we get: $\frac{C P}{P B}=\frac{C R}{R D} \quad \cdots(2)$ Hence, from $(1)$ and $(2)$, we have : $\frac{C Q}{Q A}=\frac{C R}{R...
Read More →If the difference between the circumference and radius of a circle is 37 cm,
Question: If the difference between the circumference and radius of a circle is $37 \mathrm{~cm}$, then using $\pi=\frac{22}{7}$, the circumference (in $\mathrm{cm}$ ) of the circle is (a) 154(b) 44(c) 14(d) 7 Solution: We know that circumference; $\mathrm{C}$ of the circle with radius $r$ is equal to $2 \pi r$. We have given difference between circumference and radius of the circle that is $37 \mathrm{~cm}$. $\therefore C-r=2 \pi r-r$ $\therefore(2 \pi-1) r=37$ Substituting $\pi=\frac{22}{7}$ w...
Read More →In △ABC, M and N are points on AB and AC
Question: In △ABC, M and N are points on AB and AC respectively such that BM = CN. If B = C then show that MN∥BC. Solution: In △ABC, B = CAB = AC (Sides opposite to equal angle are equal)Subtracting BM from both sides, we getAB BM = AC BM⇒AB BM = AC CN (∵BM = CN)⇒AM = ANAMN = ANM (Angles opposite to equal sides are equal)Now, In △ABC,A + B + C = 180o .....(1) (Angle Sum Property of triangle)Again In In △AMN,A +AMN + ANM = 180o ......(2) (Angle Sum Property of triangle)From (1) and (2), we getB +...
Read More →The following data were obtained during the first order thermal decomposition of
Question: The following data were obtained during the first order thermal decomposition of SO2Cl2at a constant volume. $\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ Calculate the rate of the reaction when total pressure is 0.65 atm. Solution: The thermal decomposition of SO2Cl2at a constant volume is represented by the following equation. After time, $t$, total pressure, $\mathrm{P}_{t}=\left(\mathrm{P}_{0}-p\right)+p+p$ ...
Read More →In the adjoining figure, ABCD is a trapezium in which CD ∥ AB and its diagonals intersect at O.
Question: In the adjoining figure,ABCDis a trapezium in whichCD∥ABand its diagonals intersect atO. IfAO= (2x+ 1) cm,OC= (5x 7) cm,DO= (7x 5) cm andOB= (7x+ 1) cm, find the value ofx. Solution: In trapezium $\mathrm{ABCD}, A B \| C D$ and the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ intersect at $\mathrm{O}$. Therefore, $\frac{A O}{O C}=\frac{B O}{O D}$ $\Rightarrow \frac{2 x+1}{5 x-7}=\frac{7 x+1}{7 x-5}$ $\Rightarrow(5 x-7)(7 x+1)=(7 x-5)(2 x+1)$ $\Rightarrow 35 x^{2}+5 x-49 x-7=14 x^{2}-10 x+...
Read More →If the circumference and the area of a circle
Question: If the circumference and the area of a circle are numerically equal, then diameter of the circle is (a) $\frac{\pi}{2}$ (b) $2 \pi$ (c) 2(d) 4 Solution: We have given that circumference and area of a circle are numerically equal. Let it bex. Letrbe the radius of the circle, therefore, circumference of the circle isand area of the circle will be. Therefore, from the given condition we have, $2 \pi r=x$.......(1) $\pi r^{2}=x \ldots \ldots(2)$ Therefore, from equation (1) get $r=\frac{x}...
Read More →Show that the line segment which joins the midpoints
Question: Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides. Solution: Let the trapeziumbeABCDwithEandFas the mid points ofADandBC,respectively.ProduceADandBCto meet atP. $\ln \triangle P A B, D C \| A B$ Applying Thales' theorem, we get: $\frac{P D}{D A}=\frac{P C}{C B}$ Now, $E$ and $F$ are the midpoints of $A D$ and $B C$, respectively. $\Rightarrow \frac{P D}{2 D E}=\frac{P C}{2 C F}$ $\Rightarrow \frac{P D}{D E}=\frac...
Read More →For the decomposition of azoisopropane to hexane and nitrogen at 543 K,
Question: For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. Calculate the rate constant. Solution: The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation. After time, $t$, total pressure, $\mathrm{P}_{t}=\left(\mathrm{P}_{0}-p\right)+p+p$ $\Rightarrow \mathrm{P}_{t}=\mathrm{P}_{0}+p$ $\Rightarrow p=\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{0}$ Therefore, $\mathrm{P}_{\mathrm{o}}-p=\mathrm{P}_{\m...
Read More →An arc subtends an angle of 90° at the centre of
Question: An arc subtends an angle of 90 at the centre of the circle of the radius 14 cm. Write the area of minor sector thus formed in terms of . Solution: We have given an angle subtended by an arc at the centre of the circle and radius of the circle. $r=14 \mathrm{~cm}$ $\theta=90^{\circ}$ Now we will find the area of the minor sector. Area of the minor sector $=\frac{\theta}{360} \times \pi r^{2}$ Substituting the values we get, Area of the minor sector $=\frac{90}{360} \times \pi \times 14^...
Read More →M is a point on the side BC of a parallelogram ABCD.
Question: M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that (i) $\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$ (ii) $\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}$ Solution: Given: ABCD is a parallelogramTo prove: (i) $\frac{\mathrm{DM}}{\mathrm{MN}}=\frac{\mathrm{DC}}{\mathrm{BN}}$ (ii) $\frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{AN}}{\mathrm{DC}}$ Proof: In $\triangle D M C$ and $\triangle N M B$ $\a...
Read More →A first order reaction takes 40 min for 30% decomposition.
Question: A first order reaction takes 40 min for 30% decomposition. Calculatet1/2. Solution: For a first order reaction, $t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ $k=\frac{2.303}{40 \mathrm{~min}} \log \frac{100}{100-30}$ $=\frac{2.303}{40 \mathrm{~min}} \log \frac{10}{7}$ $=8.918 \times 10^{-3} \mathrm{~min}^{-1}$ Therefore,t1/2of the decomposition reaction is $t_{1 / 2}=\frac{0.693}{k}$ $=\frac{0.693}{8.918 \times 10^{-3}} \mathrm{~min}$ = 77.7 min (approximately)...
Read More →If the diameter of a semi-circular protractor is 14 cm,
Question: If the diameter of a semi-circular protractor is 14 cm, then find its perimeter. Solution: LetABbe the diameter of the semi-circular protractor. So, $A B=14 \mathrm{~cm}$ We know that perimeter of the semicircle $=\frac{1}{2}(2 \pi r)+2 r$ We have given the diameter of the protractor. Therefore, radius of the protractor $=\frac{14}{2}$ So, radius of the protractor Substituting the value ofrin equation (1) we get, Perimeter of the semi-circular protractor $=\frac{1}{2}(2 \pi \times 7)+2...
Read More →For a first order reaction,
Question: For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. Solution: For a first order reaction, the time required for 99% completionis $t_{1}=\frac{2.303}{k} \log \frac{100}{100-99}$ $=\frac{2.303}{k} \log 100$ $=2 \times \frac{2.303}{k}$ For a first order reaction, the time required for 90% completion is $t_{2}=\frac{2.303}{k} \log \frac{100}{100-90}$ $=\frac{2.303}{k} \log 10$ $=\frac{2.303}{k}$ Therefore,...
Read More →Solve this
Question: In a $\Delta A B C, A D$ is the bisector or $\angle A$. (i) IfAB= 6.4 cm,AC= 8 cm andBD =5.6 cm, findDC.(ii) IfAB= 10 cm,AC= 14 cm andBC =6 cm, findBDandDC.(iii) IfAB= 5.6 cm,BD= 3.8 cm andBC =6 cm, findAC.(iv) IfAB= 5.6 cm,AC= 4 cm andDC =3 cm, findBC. Solution: (i) It is given that $\mathrm{AD}$ bisects $\angle \mathrm{A}$. Applying angle - bisector theorem in $\triangle \mathrm{ABC}$, we get: $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$ $\Rightarrow \frac{5.6}{\...
Read More →During nuclear explosion, one of the products is
Question: During nuclear explosion, one of the products is90Sr with half-life of 28.1 years. If 1g of90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically. Solution: Here, $k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{28.1} \mathrm{y}^{-1}$ It isknown that, $t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ $\Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]}$ $\Rightarro...
Read More →If the adjoining figure is a sector of a circle of radius 10.5 cm,
Question: If the adjoining figure is a sector of a circle of radius $10.5 \mathrm{~cm}$, what is the perimeter of the sector? (Take $\pi=22 / 7$ ) Solution: Given figure is a quadrant of a circle. We have given radius of sector that is 10.5 cm. Arc AB subtended an angle of 60 at the centre of the circle. Perimeter of the sector $=\frac{\theta}{360} \times 2 \pi r+2 r$ Substituting the values we get, Perimeter of the sector $=\frac{60}{360} \times 2 \pi \times 10.5+2 \times 10.5$ Now we will simp...
Read More →The rate constant for a first order reaction is
Question: The rate constant for a first order reaction is 60 s1. How much time will it take to reduce the initial concentration of the reactant to its 1/16thvalue? Solution: It isknown that, $t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ $=\frac{2.303}{60 \mathrm{~s}^{-1}} \log 16$ $=4.6 \times 10^{-2} \mathrm{~s}$ (approximately) Hence, the required time is 4.6 102s....
Read More →Write the formula for the area of a segment
Question: Write the formula for the area of a segment in a circle of radius r given that the sector angle is(in degrees). Solution: In this figure, centre of the circle is $O$, radius $O A=r$ and $\angle A O B=\theta$ We are going to find the area of the segmentAXB. Area of the segment $A X B=$ Area of the sector $O A X B$ - Area of $\triangle A O B$ We know that area of the sector $O A X B=\frac{\theta}{360} \times \pi r^{2}$ We also know that area of $\triangle A O B=r^{2} \sin \frac{\theta}{2...
Read More →The experimental data for decomposition of
Question: The experimental data for decomposition of $\left[2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\right]$ in gas phase at 318K are given below: (i)Plot [N2O5] againstt. (ii)Find the half-life period for the reaction. (iii)Draw a graph between log [N2O5] andt. (iv)What is the rate law? (v)Calculate the rate constant. (vi)Calculate the half-life period fromkand compare it with (ii). Solution: (i) (ii)Time corresponding to the concentration, $\frac{1.630 ...
Read More →Write the formula for the area of a sector
Question: Write the formula for the area of a sector of angle(in degrees) of a circle of radius r. Solution: Letrbe the radius of the circle and anglesubtended at the centre of the circle. Area of the sector of the circle $=\frac{\theta}{360} \times \pi r^{2}$ Therefore, area of the sector is $\frac{\theta}{360} \times \pi r^{2}$...
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