eSaral - JEE | NEET | Class 8-10 Exam Prep App

Two circular pieces of equal radii and maximum area,

Question: Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm 7 cm. Find the area of the remaining card board. (Use = 22/7). Solution: We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard. Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm Now, Area of remaining...

In the following figure a square OABC is inscribed in a quadrant OPBQ of a circle.

Question: In the following figure a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region. Solution: Construction: Join OB In right triangle AOB $\mathrm{OB}^{2}=\mathrm{OA}^{2}+\mathrm{AB}^{2}$ $=21^{2}+21^{2}$ $=441+441$ $=882$ $\therefore \mathrm{OB}^{2}=882$ Area of the shaded region = Area of quadrant OPBQ Area of Square OABC $=\frac{1}{4} \pi(\mathrm{OB})^{2}-(\mathrm{OA})^{2}$ $=\frac{1}{4} \times \frac{22}{7} \times 882-441$ $=693-441$...

For the reaction:

Question: For the reaction: 2A + B A2B the rate =k[A][B]2withk= 2.0 106mol2L2s1. Calculate the initial rate of the reaction when [A] = 0.1 mol L1, [B] = 0.2 mol L1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L1. Solution: The initial rate of the reactionis Rate =k[A][B]2 = (2.0 106mol2L2s1) (0.1 mol L1) (0.2 mol L1)2 = 8.0 109mol2L2s1 When[A] is reduced from 0.1 mol L1to 0.06 mol1, the concentration of A reacted = (0.1 0.06) mol L1= 0.04 mol L1 Therefore, concentration of $B...

From the rate expression for the following reactions,

Question: From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i)3 NO(g) N2O(g) Rate =k[NO]2 (ii)H2O2(aq) + 3 I(aq) + 2 H+ 2 H2O (l) +Rate =k[H2O2][I] (iii)CH3CHO(g) CH4(g) + CO(g) Rate =k[CH3CHO]3/2 (iv)C2H5Cl(g) C2H4(g) + HCl(g) Rate =k[C2H5Cl] Solution: (i)Given rate =k[NO]2 Therefore, order of the reaction = 2 Dimension of $k=\frac{\text { Rate }}{[\mathrm{NO}]^{2}}$ $=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}...

In the following figure, shows a kite in which BCD is the shape of

Question: In the following figure, shows a kite in whichBCDis the shape of a quadrant of a circle of radius 42 cm.ABCDis a square and ΔCEFis an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region. Solution: We will find the area of the shaded region as shown below, Area of the shaded region = area of quadrant + area of isosceles triangle ..(1) $\therefore$ Area of shaded region $=\frac{90}{360} \times \pi \times 42^{2}+\frac{1}{2} \times 6 \times 6...

If the distance between the points A(4, p) and B(1, 0) is 5, then

Question: If the distance between the pointsA(4,p) andB(1, 0) is 5, then(a)p= 4 only(b)p= 4 only(c)p= 4 only(d)p= 0 Solution: (c)p= 4 onlyThe given points areA(4,p) andB(1, 0) andAB= 5. Then, $\left(x_{1}=4, y_{1}=p\right)$ and $\left(x_{2}=1, y_{2}=0\right)$ Therefore, $A B=5$ $\Rightarrow \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=5$ $\Rightarrow \sqrt{(1-4)^{2}+(0-p)^{2}}=5$ $\Rightarrow(-3)^{2}+(-p)^{2}=25$ $\Rightarrow 9+p^{2}=25$ $\Rightarrow p^{2}=16$ $\Rightarrow p=...

In the following figure, the boundary of the shaded region consists

Question: In the following figure, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, find(i) the length of the boundary.(ii) the area of the shaded region. Solution: (i) We will first find the length of the boundary. Length of the boundaryperimeter of semi-circle with diameter AB + boundary of semi-circle with diameter 7 cm Length of the boundary $=(\pi \times 7+\pi \times 3....

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0).

Question: AOBCis a rectangle whose three vertices areA(0, 3),O(0, 0) andB(5, 0). Length of each of its diagonal is (a) 5 units (b) 3 units (c) $\sqrt{34}$ units (d) 4 units Solution: (c) $\sqrt{34}$ units $A(0,3), O(0,0)$ and $B(5,0)$ are the three vertices of a rectangle; let $C$ be the fourth vertex. Then, the length of the diagonal, $A B=\sqrt{(5-0)^{2}+(0-3)^{2}}$ $=\sqrt{(5)^{2}+(-3)^{2}}$ $=\sqrt{25+9}$ $=\sqrt{34}$ units Since, the diagonals of rectangle are equal . Hence, the length of i...

The area of ∆ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is

Question: The area of ∆ABCwith verticesA(3, 0),B(7, 0) andC(8, 4) is(a) 14 sq units(b) 28 sq units(c) 8 sq units(d) 6 sq units Solution: (c) 8 sq units The given points are $A(3,0), B(7,0)$ and $C(8,4)$. Here, $\left(x_{1}=3, y_{1}=0\right),\left(x_{2}=7, y_{2}=0\right)$ and $\left(x_{3}=8, y_{3}=4\right)$ Therefore, Area of $\Delta A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)]$ $=\frac...

The activation energy for the reaction

Question: The activation energy for the reaction 2HI(g)H2+I2(g) is 209.5 kJmol1at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? Solution: In the given case: Ea= 209.5 kJmol1= 209500 Jmol1 T= 581 K R= 8.314JK1mol1 Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as: $x=e^{-E_{a} / \mathrm{R} T}$ $\Rightarrow \operatorname{In} x=-\frac{E_{a}}{\mathrm{R} T}$ $\Rightarrow \...

If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then

Question: If the pointsA(1, 2),O(0, 0) andC(a,b) are collinear, then(a)a=b(b)a= 2b(c) 2a=b(d)a+b= 0 Solution: (c) 2a=bThe given points areA(1, 2),O(0, 0) andC(a,b). Here, $\left(x_{1}=1, y_{1}=2\right),\left(x_{2}=0, y_{2}=0\right)$ and $\left(x_{3}=a, y_{3}=b\right)$ PointsA, OandC are collinear. $\Rightarrow x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$ $\Rightarrow 1(0-b)+0(b-2)+a(2-0)=0$ $\Rightarrow-b+2 a=0$ $\Rightarrow 2 a=b$...

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K.

Question: The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. CalculateEa. Solution: It is given thatT1= 298 K T2= (298 + 10) K = 308 K We also know that the rate of the reaction doubles when temperature is increased by 10. Therefore, let us take the value ofk1=kand that ofk2= 2k Also,R= 8.314 J K1mol1 Now, substituting these values in the equation: $\log \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\righ...

In the following figure, ABC is an equilateral triangle of side 8 cm.

Question: In the following figure, $A B C$ is an equilateral triangle of side $8 \mathrm{~cm} . A, B$ and $C$ are the centres of circular arcs of radius $4 \mathrm{~cm}$. Find the area of the shaded region correct upto 2 decimal places. (Take $\pi=3.142$ and $\sqrt{3}=1.732$ ). Solution: Area of the shaded region can be calculated as shown below, Area of the shaded region = Area of equilateral triangle 3area of circular arc $\therefore$ Area of the shaded region $=\frac{\sqrt{3}}{4} \times 8 \ti...

What will be the effect of temperature on rate constant?

Question: What will be the effect of temperature on rate constant? Solution: The rate constantof a reaction is nearly doubled with a 10rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation, $k=\mathrm{Ae}^{-E \mathrm{a} / R T}$ Where, A is the Arrheniusfactor or the frequency factor Tis the temperature Ris the gas constant Eais the activation energy...

If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

Question: If the pointsA(2, 3),B(5,k) andC(6, 7) are collinear, then (a) $k=4$ (b) $k=6$ (c) $k=\frac{-3}{2}$ (d) $k=\frac{11}{4}$ Solution: (b)k= 6The given points areA(2, 3),B(5,k) andC(6, 7). Here, $\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=5, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=7\right)$ PointsA,BandCare collinear. Then, $x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$ $\Rightarrow 2(k-7)+5(7-3)+6(3-k)=0$ $\Rightarrow 2 k-14+20+18-6 k=0$ $\R...

Time required to decompose

Question: Time required to decompose SO2Cl2to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. Solution: We know that for a 1storder reaction, $t_{1 / 2}=\frac{0.693}{k}$ It is given thatt1/2= 60 min $\therefore k=\frac{0.693}{t_{1 / 2}}$ $=\frac{0.693}{60}$ $=0.01155 \mathrm{~min}^{-1}$ $=1.155 \mathrm{~min}^{-1}$ Or $k=1.925 \times 10^{-4} \mathrm{~s}^{-1}$...

In the following figure, ABCD is a rectangle with AB = 14 cm and BC = 7 cm.

Question: In the following figure,ABCDis a rectangle withAB= 14 cm andBC= 7 cm. TakingDC,BCandADas diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region. Solution: Area of the shaded region can be calculated as shown below, Area of the shaded region = Area of rectangle area of the semi-circle with diameter DC triangle + 2area of two semicircles with diameters AD and BC $\therefore$ Area of the shaded region $=7 \times 14-\frac{\pi \times 7^{2}}{2}+2 \...

A first order reaction has a rate constant

Question: A first order reaction has a rate constant 1.15 103s1. How long will 5 g of this reactant take to reduce to 3 g? Solution: From the question, we can writedown the following information: Initial amount = 5 g Final concentration = 3 g Rate constant = 1.15 103s1 We know that for a 1storder reaction, $t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ $=\frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3}$ $=\frac{2.303}{1.15 \times 10^{-3}} \times 0.2219$ = 444.38 s = 444 s (appr...

In the following figure, two circles with centres A and B touch each other at the point C.

Question: In the following figure, two circles with centresAandBtouch each other at the pointC. IfAC= 8 cm andAB= 3 cm, find the area of the shaded region. Solution: Area of the shaded region can be calculated as shown below, Area of the shaded region = Area of circle with radius AC area of circle with radius BC We have given radius of the outer circle that is 8 cm but we dont know the radius of the inner circle. We can calculate the radius of the inner circle as shown below, $B C=A C-A B$ $\the...

The conversion of molecules X to Y follows second order kinetics.

Question: The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y? Solution: The reaction X Y follows second order kinetics. Therefore, the rate equation for this reaction will be: Rate =k[X]2(1) Let [X] =amol L1, then equation (1) can be written as: Rate1=k.(a)2 =ka2 If the concentration of X is increased to three times, then [X] = 3amol L1 Now, the rate equation will be: Rate =k(3a)2 = 9(k...

In the following figure,

Question: In the following figure, ABCD is a square of side 2a, Find the ratio between(i) the circumferences(ii) the areas of the in circle and the circum-circle of the square. Solution: We have a square $\mathrm{ABCD}$ having $\mathrm{AB}=2 a$. From the given diagram we can observe that, Radius of incircle $\left(r_{1}\right)=a$ Radius of circumcircle $\left(r_{2}\right)=\sqrt{2} a$ (i) We have to find the ratio of the circumferences of the two circles. So the required ratio is, $=\frac{\text {...

For a reaction, A + B → Product; the rate law is given by

Question: For a reaction, $\mathrm{A}+\mathrm{B} \rightarrow$ Product; the rate law is given by, $r=k[\mathrm{~A}]^{1 / 2}[\mathrm{~B}]^{2}$. What is the order of the reaction? Solution: The order of the reaction $=\frac{1}{2}+2$ $=2 \frac{1}{2}$ = 2.5...

In a reaction, 2A → Products,

Question: In areaction, 2AProducts, the concentration of A decreases from 0.5 mol L1to 0.4 mol L1in 10 minutes. Calculate the rate during this interval? Solution: Average rate $=-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta t}$ $=-\frac{1}{2} \frac{[\mathrm{A}]_{2}-[\mathrm{A}]_{1}}{t_{2}-t_{1}}$ $=-\frac{1}{2} \frac{0.4-0.5}{10}$ $=-\frac{1}{2} \frac{-0.1}{10}$ = 0.005 mol L1min1 = 5103M min1...

The points P(0, 6), Q(−5, 3) and R(3, 1) are the vertices of a triangle, which is

Question: The pointsP(0, 6),Q(5, 3) andR(3, 1) are the vertices of a triangle, which is(a) equilateral(b) isosceles(c) scalene(d) right-angled Solution: (d) right-angledLetP(0, 6),Q(5, 3) andR(3, 1) be the given points. Then, $P Q=\sqrt{(-5-0)^{2}+(3-6)^{2}}$ $=\sqrt{(-5)^{2}+(-3)^{2}}$ $=\sqrt{25+9}$ $=\sqrt{34}$ units $Q R=\sqrt{(3+5)^{2}+(1-3)^{2}}$ $=\sqrt{(8)^{2}+(-2)^{2}}$ $=\sqrt{64+4}$ $=\sqrt{68}$ $=2 \sqrt{17}$ units $P R=\sqrt{(3-0)^{2}+(1-6)^{2}}$ $=\sqrt{(3)^{2}+(-5)^{2}}$ $=\sqrt{9...

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes.

Question: For the reaction RP, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. Solution: Average rate of reaction $=-\frac{\Delta[\mathrm{R}]}{\Delta t}$ $=-\frac{[\mathrm{R}]_{2}-[\mathrm{R}]_{1}}{t_{2}-t_{1}}$ $=-\frac{0.02-0.03}{25} \mathrm{M} \mathrm{min}^{-1}$ $=-\frac{-0.01}{25} \mathrm{M} \mathrm{min}^{-1}$ $=4 \times 10^{-4} \mathrm{M} \mathrm{min}^{-1}$ $=\frac{4 \times 1...

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