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

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.

Question: The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. Solution: From Arrhenius equation, we obtain $\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$ It is given that, $k_{2}=4 k_{1}$ $T_{1}=293 \mathrm{~K}$ $T_{2}=313 \mathrm{~K}$ Therefore, $\log \frac{4 k_{1}}{k_{2}}=\frac{E_{a}}{2.303 \times 8.314}\lef...

If the circumference of a circle increases from 4π to 8π, then its area is

Question: If the circumference of a circle increases from 4 to 8, then its area is (a) halved(b) doubled(c) tripled(d) quadrupled Solution: Let the circumference $C=4 \pi$ $\therefore 2 \pi r=4 \pi$ $\therefore r=2$ Therefore, area of the circle when radius of the circle is 2 can be calculated as below, $\pi r^{2}=\frac{22}{7} \times 4$ ......(1) Now when circumference is, then the radius of the circle is calculated as below, $\therefore 2 \pi r=8 \pi$ $\therefore r=4$ Therefore, area of the cir...

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K.

Question: The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value ofAis 4 1010s1. Calculatekat 318 K andEa. Solution: For a first order reaction, $t=\frac{2.303}{k} \log \frac{a}{a-x}$ At $298 \mathrm{~K}, t=\frac{2.303}{k} \log \frac{100}{90}$ $=\frac{0.1054}{k}$ At $308 \mathrm{~K}, t^{\prime}=\frac{2.303}{k^{\prime}} \log \frac{100}{75}$ $=\frac{2.2877}{k^{\prime}}$ According to the question, $t=t^{\prime}...

In the given figure

Question: In the given figure $\triangle O D C \sim \triangle O B A, \angle B O C=115^{\circ}$ and $\angle C D O=70^{\circ}$. Find (i) DOC(ii) DCO(iii) OAB(iv) OBA. Solution: (i)It is given that DB is a straight line.Therefore, $\angle D O C+\angle C O B=180^{\circ}$ $\angle D O C=180^{\circ}-115^{\circ}=65^{\circ}$ (ii) In $\triangle D O C$, we have: $\angle O D C+\angle D C O+\angle D O C=180^{\circ}$ Therefore, $70^{\circ}+\angle D C O+65^{\circ}=180^{\circ}$ $\Rightarrow \angle D C O=180-70-...

The area of a circle is 220 cm2.

Question: The area of a circle is $220 \mathrm{~cm}^{2}$. The area of ta square inscribed in it is (a) $49 \mathrm{~cm}^{2}$ (b) $70 \mathrm{~cm}^{2}$ (c) $140 \mathrm{~cm}^{2}$ (d) $150 \mathrm{~cm}^{2}$ Solution: Let BD be the diameter and diagonal of the circle and the square respectively. We know that area of the circle $=\pi r^{2}$ Area of the circle $=\pi r^{2}$ $\therefore 220=\frac{22}{7} \times r^{2}$ Multiplying both sides of the equation by 7 we get, $220 \times 7=22 \times r^{2}$ Div...

The decomposition of A into product has value of

Question: The decomposition of A into product has value ofkas 4.5 103s1at 10C and energy of activation 60 kJ mol1. At what temperature wouldkbe 1.5 104s1? Solution: From Arrhenius equation, we obtain $\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$ Also,k1= 4.5 103s1 T1= 273 + 10 = 283 K k2= 1.5 104s1 Ea= 60 kJ mol1= 6.0 104J mol1 Then, $\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}=\frac{6.0 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}}{2....

Find the pair of similar triangles among the given pairs.

Question: Find the pair of similar triangles among the given pairs. State the similarity criterion and write the similarity relation in symbolic form. Solution: (i)We have: $\angle B A C=\angle P Q R=50^{\circ}$ $\angle A B C=\angle Q P R=60^{\circ}$ $\angle A C B=\angle P R Q=70^{\circ}$ Therefore, by AAA similarity theorem, $\triangle A B C \sim Q P R$ (ii)We have: $\frac{A B}{D F}=\frac{3}{6}=\frac{1}{2}$ and $\frac{B C}{D E}=\frac{4.5}{9}=\frac{1}{2}$ But, $\angle A B C \neq \angle E D F$ (I...

The decomposition of A into product has value of

Question: The decomposition of A into product has value ofkas 4.5 103s1at 10C and energy of activation 60 kJ mol1. At what temperature wouldkbe 1.5 104s1? Solution: gfdh...

The perimeter of a triangle is 30 cm

Question: The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is (a) $70 \mathrm{~cm}^{2}$ (b) $140 \mathrm{~cm}^{2}$ (c) $210 \mathrm{~cm}^{2}$ (d) $420 \mathrm{~cm}^{2}$ Solution: We have to find the area of the given triangle. Perimeter of triangle is 30 cm. Let the radius of the circle ber. We have, Circumference of incircle $=88$ $2 \pi r=88$ $r=14$ Therefore, $a r(\Delta \mathrm{ABC})=a r(\Delta \mathrm{OAB})+a r(\Delta \mathrm{OBC}...

The area of the largest triangle that can be inscribed

Question: The area of the largest triangle that can be inscribed in a semi-circle of radius r, is (a) $r^{2}$ (b) $2 r^{2}$ (c) $r^{3}$ (d) $2 r^{3}$ Solution: The triangle with the largest area will be symmetrical as shown in the figure. Let the radius of the circle ber. Hence, $a r(\Delta \mathrm{ABC})=\frac{1}{2}(r)(2 r)$ $=r^{2} s q$, unit Therefore the answer is (a)....

The decomposition of A into product has value of

Question: The decomposition of A into product has value ofkas 4.5 103s1at 10C and energy of activation 60 kJ mol1. At what temperature wouldkbe 1.5 104s1? Solution: From Arrhenius equation, we obtain $\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$ Also,k1= 4.5 103s1 T1= 273 + 10 = 283 K k2= 1.5 104s1 Ea= 60 kJ mol1= 6.0 104J mol1 Then, $\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}=\frac{6.0 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}}{2....

The area of incircle of an equilateral triangle is 154 cm2

Question: The area of incircle of an equilateral triangle is $154 \mathrm{~cm}^{2}$. The perimeter of the triangle is (a) 71.5 cm(b) 71.7 cm(c) 72.3 cm(d) 72.7 cm Solution: Area of incircle of equilateral triangle is $154 \mathrm{~cm}^{2}$ We have to find the perimeter of the triangle. So we will use area to get, Area of incircle $=154$ $\pi r^{2}=154$ $r=\sqrt{\frac{154}{\pi}} \mathrm{cm}$ As triangle is equilateral so, $\angle \mathrm{OCM}=30^{\circ}$ So, $\tan 30^{\circ}=\frac{r}{\mathrm{MC}}...

The area of the incircle of an equilateral triangle of side 42 cm is

Question: The area of the incircle of an equilateral triangle of side 42 cm is (a) $22 \sqrt{3} \mathrm{~cm}^{2}$ (b) $231 \mathrm{~cm}^{2}$ (c) $462 \mathrm{~cm}^{2}$ (d) $924 \mathrm{~cm}^{2}$ Solution: Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Also, let O be the centre andr be the radius of itsincircle. AB, BC and CA are tangents to the circle at M, N and P. OM = ON = OP =r Area of ΔABC = Area (ΔOAB) + Area (ΔOBC) + Area (ΔOCA) $\Rightarrow \frac{\sqrt{3}}{4}(42)^{2}...

The circumference of a circle is 100 m.

Question: The circumference of a circle is 100 m. The side of a square inscribed in the circle is (a) $50 \sqrt{2}$ (b) $\frac{50}{\pi}$ (c) $\frac{50 \sqrt{2}}{\pi}$ (d) $\frac{100 \sqrt{2}}{\pi}$ Solution: We have given the circumference of the circle that is 100 cm. Ifdis the diameter of the circle, then its circumference will be. $\therefore \pi d=100$ $\therefore d=\frac{100}{\pi}$ We obtained diameter of the circle. Look at the figure, diameter of the circle is also the diagonal of the squ...

The ratio of the outer and inner perimeters of a

Question: The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is(a) 55 m(b) 110 m(c) 220 m(d) 230 m Solution: Let OA =rbe the radius of the inner circle and OB =rbe the radius of the outer circle. Therefore, circumference of the inner circle $=2 \pi r$ and circumference of the outer circle $=2 \pi r^{\prime}$ Here we have to find the diameter of the inner circle. We have given the ratio of outer and inner perim...

The rate constant for the first order decomposition of

Question: The rate constant for the first order decomposition of H2O2is given by the following equation: logk= 14.34 1.25 104K/T CalculateEafor this reaction and at what temperature will its half-period be 256 minutes? Solution: Arrhenius equation is given by, $k=\mathrm{Ae}^{-E_{\alpha} / \mathrm{R} T}$ $\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$ $\Rightarrow \ln k=\log \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$ $\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{RT}}$ ...

The rate constant for the first order decomposition of

Question: The rate constant for the first order decomposition of H2O2is given by the following equation: logk= 14.34 1.25 104K/T CalculateEafor this reaction and at what temperature will its half-period be 256 minutes? Solution: Arrhenius equation is given by, $k=\mathrm{Ae}^{-E_{\alpha} / \mathrm{R} T}$ $\Rightarrow \ln k=\ln \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$ $\Rightarrow \ln k=\log \mathrm{A}-\frac{E_{a}}{\mathrm{R} T}$ $\Rightarrow \log k=\log \mathrm{A}-\frac{E_{a}}{2.303 \mathrm{RT}}$ ...

In ∆ABC, the bisector of ∠B meets AC at D. A line PQ ∥ AC meets AB, BC and BD at P,

Question: In ∆ABC, the bisector of BmeetsACatD. A linePQ∥ACmeetsAB,BCandBDatP, QandRrespectively.Show thatPRBQ=QRBP. Solution: In triangle BQP, BR bisects angle B.Applying angle bisector theorem, we get: $\frac{Q R}{P R}=\frac{B Q}{B P}$ $\Rightarrow B P \times Q R=B Q \times P R$ This completes the proof....

In the adjoining figure, ABC is a triangle in which AB = AC.

Question: In the adjoining figure,ABCis a triangle in whichAB=AC. IfDandEare points onABandACrespectively such thatAD=AE, show that the pointsB, C, EandDare concyclic. Solution: Given:AD = AE ...(i)AB = AC ...(ii)Subtracting AD from both sides, we get:⇒⇒AB-AD = AC-AD⇒⇒AB-AD = AC-AE (Since,AD = AE)⇒⇒BD = EC ...(iii)Dividing equation (i) by equation (iii), we get: $\frac{A D}{D B}=\frac{A E}{E C}$ Applying the converse of Thales' theorem, $D E \| B C$ $\Rightarrow \angle D E C+\angle E C B=180^{\c...

ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that

Question: $A B C D$ is a parallelogram in which $P$ is the midpoint of $D C$ and $Q$ is a point on $A C$ such that $C Q=\frac{1}{4} A C .$ If $P Q$ produced meets $B C$ at $R$, prove that $R$ is the midpoint $B C$. Solution: Join DB.We know that the diagonals of a parallelogram bisect each other.Therefore $C S=\frac{1}{2} A C \quad \ldots$ (i) Also, it is given that $C Q=\frac{1}{4} A C$ ...(ii) Dividing equation (ii) by (i), we get: $\frac{C Q}{C S}=\frac{\frac{1}{4} A C}{\frac{1}{2} A C}$ or, ...

The radius of a wheel is 0.25 m.

Question: The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be(a) 2800 (b) 4000(c) 5500(d) 7000 Solution: We have given the radius of the wheel that is 0.25 cm. We know that distance covered by the wheel in one revolution $=\frac{\text { Distance moved }}{\text { Number of revolutions }}$. Distance covered by the wheel in one revolution is equal to the circumference of the wheel. $2 \pi r=\frac{\text { Distance moved }}{\text { Number of r...

The decomposition of hydrocarbon follows the equation

Question: The decomposition of hydrocarbon follows the equation k= (4.5 1011s1) e28000K/T CalculateEa. Solution: The given equation is k= (4.5 1011s1) e28000K/T(i) Arrhenius equation is given by, $k=\mathrm{Ae}^{-E_{\alpha} / \mathrm{R} T}$ From equation (i) and (ii), we obtain $\frac{E_{a}}{\mathrm{R} T}=\frac{28000 \mathrm{~K}}{T}$ $\Rightarrow E_{a}=\mathrm{R} \times 28000 \mathrm{~K}$ = 8.314 J K1mol1 28000 K = 232792 J mol1 = 232.792 kJ mol1...

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law,

Question: Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, witht1/2= 3.00 hours. What fraction of sample of sucrose remains after 8 hours? Solution: For a first order reaction, $k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ It is given that,t1/2= 3.00 hours Therefore, $k=\frac{0.693}{t_{1 / 2}}$ $=\frac{0.693}{3} \mathrm{~h}^{-1}$ = 0.231 h1 Then, $0.231 \mathrm{~h}^{-1}=\frac{2.303}{8 \mathrm{~h}} \log \frac{[\mathrm{R}]_{0}}{[...

A circular park has a path of uniform width around it.

Question: A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is(a) 20 m(b) 21 m(c) 22 m(d) 24 m Solution: Let OA =rbe the radius of the inner circle and OB =rbe the radius of the outer circle. Therefore, circumference of the inner circle $=2 \pi r$ and circumference of the outer circle $=2 \pi r^{\prime}$ Here we have to find the width of the circular park that is we have to find $r^{\prime}-r$...

Consider a certain reaction A → Products

Question: Consider a certain reaction A Products withk= 2.0 102s1. Calculate the concentration ofAremaining after 100 s if the initial concentration ofAis 1.0 mol L1. Solution: k= 2.0 102s1 T= 100 s [A]o= 1.0 moL1 Sincethe unit ofkis s1, the given reaction is a first order reaction. Therefore, $k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}$ $\Rightarrow 2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}} \log \frac{1.0}{[\mathrm{~A}]}$ $\Rightarrow 2.0 \times 10^{-2...

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