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

In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its side AB,

Question: In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its side AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB =xcm, BC = 7 cm, CR = 3 cm and AS = 5 cm , findx Solution: We know that tangent segments to a circle from the same external point are congruent.Now, we haveCR = CQ, AS = AP and BQ = BPNow, BC = 7 cm⇒ CQ + BQ = 7⇒ BQ = 7 CQ⇒ BQ = 7 3 [∵ CQ = CR = 3]⇒ BQ = 4 cmAgain, AB = AP + PB= AP + BQ= 5 + 4 [∵ AS = AP = 5]= 9 cmHence, ...

Prove that in a triangle if the square of one side

Question: Prove that in a triangle if the square of one side is equal to the sum of the squares of the other two side then the angle opposite to the first side is a right angle. Solution: Given: if square of one side is equal to the sum of the squares of the other two sides, then we have to prove that the angle opposite the first side is a right angle. Here, we are given a triangle ABC with. We need to prove that. Now, we construct a triangle PQR right angled at Q such thatand. We have the follo...

In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F.

Question: In the given figure,ADandAEare the tangents to a circle with centreOandBCtouches the circle atF.IfAE= 5 cm, then perimeter of ∆ABCis (a) 15 cm(b) 10 cm(c) 22.5 cm(d) 20 cm Solution: (b) 10 cmSince the tangents from an external point are equal, we have: $A D=A E, C D=C F, B E=B F$ Perimeter of $\Delta A B C=A C+A B+C B$ $=(A D-C D)+(C F+B F)+(A E-B E)$ $=(A D-C F)+(C F+B F)+(A E-B F)$ $=A D+A E$ $=2 A E$ $=2 \times 5$ $=10 \mathrm{~cm}$...

Two identical strings X and Z made of same material

Question: Two identical strings $X$ and $Z$ made of same material have tension $T_{X}$ and $T_{Z}$ in them. If their fundamental frequencies are $450 \mathrm{~Hz}$ and $300 \mathrm{~Hz}$, respectively, then the ratio $T_{X} / T_{Z}$ is:$2.25$$0.44$$1.25$$1.5$Correct Option: 1 Solution: (1) Using $f=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$, where, $T=$ tension and $\mu=\frac{\text { mass }}{\text { length }}$ $f_{x}=\frac{1}{2 \ell} \sqrt{\frac{T_{x}}{\mu}}$ and $f_{z}=\frac{1}{2 \ell} \sqrt{\frac{...

Re-arrange suitably and find the sum in each of the following:

Question: Re-arrange suitably and find the sum in each of the following: (i) $\frac{11}{12}+\frac{-17}{3}+\frac{11}{2}+\frac{-25}{2}$ (ii) $\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$ (iii) $\frac{3}{5}+\frac{7}{3}+\frac{9}{5}+\frac{-13}{15}+\frac{-7}{3}$ (iv) $\frac{4}{13}+\frac{-5}{8}+\frac{-8}{13}+\frac{9}{13}$ (v) $\frac{2}{3}+\frac{-4}{5}+\frac{1}{3}+\frac{2}{5}$ (vi) $\frac{1}{8}+\frac{5}{12}+\frac{2}{7}+\frac{7}{12}+\frac{9}{7}+\frac{-5}{16}$ Solution: (i) $\left(\frac{11}{2}+\f...

Prove that :

Question: $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+s \operatorname{ec} \theta \operatorname{cosec} \theta$ Solution: Here we have to prove that $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$ Here we take the LHS and by using the trigonometric identities, we have $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}...

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 HZ.

Question: A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency $340 \mathrm{HZ}$. When fork A filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?$342 \mathrm{~Hz}$$335 \mathrm{~Hz}$$338 \mathrm{~Hz}$$345 \mathrm{~Hz}$Correct Option: , 2 Solution: (2) Given Before Filed : So answer should be $335 \mathrm{~Hz}$ or $345 \mathrm{~Hz}$. After Filed : After filed beat/sec decreases only in case of $335 \mathrm{~Hz}$....

If tangents PA and PB from a point P to a circle with centre O

Question: If tangentsPAandPBfrom a pointPto a circle with centreOare drawn, so that APB= 80, then POA= ? (a) 40(b) 50(c) 80(d) 60 Solution: (b) 50 From $\Delta O P A$ and $\Delta O P B$ $O A=O B(\mathrm{R}$ adi i of the same circle $)$ $O P(C$ ommon side $)$ $P A=P B$ (S ince tangents drawn from an external point to a circle are equal) $\therefore \Delta O P A \cong \Delta O P B$ (SSS rule) $\therefore \angle A P O=\angle B P O$ $\therefore \angle A P O=\frac{1}{2} \angle A P B=40^{\circ}$ $A$ n...

In Figure 4, two triangles ABC and DBC are on the

Question: In Figure 4, two triangles ABC and DBC are on the same base BC in which A = D = 90. If CA and BD meet each other at E, show that AE CE = BE DE. Solution: Given that, there are two trianglesABCandDBCare on the same baseBCin whichA=D= 90. IfCAandBDmeet each other atE, then we have to prove thatAE CE = BE DE The following figure is given From the above figure, we can easily see that $\triangle A B E$ and $\triangle D C E$ are similar triangles, therefore we can use the property of similar...

In Figure 3, PQ || CD and PR || CB. Prove that

Question: In Figure $3, P Q \| C D$ and $P R \| C B$. Prove that $\frac{A Q}{Q D}=\frac{A R}{R B}$. Solution: Given that: $P Q \| C D$ and $P R \| C B$, then we to prove that $\frac{A Q}{Q D}=\frac{A R}{R B}$ The following diagram is given We can easily see that, in the above figureare similar triangles, and also theare similar triangles. Now, we have the following properties of similar triangles, $\frac{A Q}{Q D}=\frac{A P}{P C}$ ................(1) $\frac{A R}{R B}=\frac{A P}{P C}$ ..............

If the angle between two radii of a circle is 130°,

Question: If the angle between two radii of a circle is 130, then the angle between the tangent at the ends of the radii is(a) 65(b) 40(c) 50(d) 90 Solution: (c) $50^{\circ}$ $\mathrm{OA}$ and $\mathrm{OB} a$ re the two $\mathrm{r}$ adii of $a$ circle with centre $\mathrm{O}$. Also, $A P$ and $B P$ are the tangents to the circle. Given, $\angle A O B=130^{\circ}$ Now, $\angle O A B=\angle O B A=90^{\circ}$ (S ince tangents drawn from an external point a re perpendicular to the radius at point of...

Using commutativity and associativity of addition of rational numbers,

Question: Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number: (i) $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$ (ii) $\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}$ (iii) $\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}$ (iv) $\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}$ Solution: (i) We have: $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$ $=\left(\frac{2}{5}+\frac{-4}{5}\right)+\lef...

The mass per unit length of a uniform wire is 0.135 g /cm.

Question: The mass per unit length of a uniform wire is $0.135 \mathrm{~g} / \mathrm{cm}$. A transverse wave of the form $y=-0.21 \sin (x+30 t)$ is produced in it, where $x$ is in meter and $\mathrm{t}$ is in second. Then, the expected value of tension in the wire is______ $x \times 10^{-2} \mathrm{~N}$. Value of $x$ is (Round-off to the nearest integer) Solution: (1215) $y=-0.21 \sin (x+30 t)$ $v=\frac{\omega}{K}=\frac{30}{1}=30 \mathrm{~m} / \mathrm{s}$ $v=\sqrt{\frac{T}{\mu}}$ $\mathrm{T}=\ma...

If sec 4A = cosec (A − 20°),

Question: If sec 4A= cosec (A 20), where 4Ais an acute angle, find the value ofA. Solution: Given that: $\sec 4 A=\operatorname{cosec}\left(A-20^{\circ}\right)$, then we have to find the value of $A$ Since it is given that the angle 4Ais acute angle, therefore we can apply the identity Therefore the given equation can be written as $\operatorname{cosec}\left(90^{\circ}-4 A\right)=\operatorname{cosec}\left(A-20^{\circ}\right)$ $\Rightarrow \quad 90^{\circ}-4 A=A-20^{\circ}$ $\Rightarrow \quad 5 A...

In the given figure, O is the centre of a circle,

Question: In the given figure,Ois the centre of a circle,PQis a chord and the tangentPTatPmakes an angle of 50 withPQ.Then, POQ= ? (a) 130(b) 100(c) 90(d) 75 Solution: (b) $100^{\circ}$ Given, $\angle Q P T=50^{\circ}$ Now, $\angle O P T=90^{\circ}$ (S ince tangents drawn from an external point a re perpendicular to the radius at point of contact) $\therefore \angle O P Q=(\angle O P T-\angle Q P T)=\left(90^{\circ}-50^{\circ}\right)=40^{\circ}$ $O P=O Q$ (Radii of the same circle) $\Rightarrow ...

A student is performing the experiment of resonance column.

Question: A student is performing the experiment of resonance column. The diameter of the column tube is $6 \mathrm{~cm}$. The frequency of the tuning fork is $504 \mathrm{~Hz}$. Speed of the sound at the given temperature is $336 \mathrm{~m} / \mathrm{s}$. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is :$13 \mathrm{~cm}$$14.8 \mathrm{~cm}$$16.6 \mathrm{~cm}$$18.4 \mathrm{~cm}$Co...

In Figure 2, ABCD is a parallelogram.

Question: In Figure 2, ABCD is a parallelogram. Find the values of x and y. Solution: The given parallelogram is We have to find the value ofxandy. We know that if two diagonals are drawn in a parallelogram, then the intersection points is the mid-point of the two diagonals. Therefore, we have two equations as follow x + y= 9 (1) x y =5 (2) Now, add the equation (1) and equation (2), we get $2 x=14$ $\Rightarrow \quad x=7$ Now, we are going to put the value ofxin equation (1), we get $7+y=9$ $\R...

Write the negative (additive inverse) of each of the following:

Question: Write the negative (additive inverse) of each of the following: (i) $\frac{-2}{5}$ (ii) $\frac{7}{-9}$ (iii) $\frac{-16}{13}$ (iv) $\frac{-5}{1}$ (v) 0 (vi) 1 (vii) 1 Solution: (i) Negative of $\frac{-2}{5}$ is $\frac{2}{5}$. (ii) Negative of $\frac{7}{-9}$ is $\frac{7}{9}$. (iii) Negative of $\frac{-16}{13}$ is $\frac{16}{13}$. (iv) Negative of $\frac{-5}{1}$ is $\frac{5}{1}$. (v) Negative value of 0 is 0 . (vi) Negative of 1 is $-1$ (vii) Negative of $-1$ is 1 ....

Find a quadratic polynomial with zeroes

Question: Find a quadratic polynomial with zeroes $3+\sqrt{2}$ and $3-\sqrt{2}$ Solution: Given that the zeroes of the quadratic polynomial are $3+\sqrt{2}$ and $3-\sqrt{2}$. We have to find the quadratic polynomial from the given zeroes. Let we assume that, $\alpha=3+\sqrt{2}$ $\beta=3-\sqrt{2}$, then $\alpha+\beta=3+\sqrt{2}+3-\sqrt{2}$ $=6$ $\alpha \beta=(3+\sqrt{2})(3-\sqrt{2})$ $=7$ Therefore the quadratic equation is given by $x^{2}-(\alpha+\beta) x+\alpha \beta=x^{2}-6 x+7$ Hence the desi...

Solve this

Question: The correct answer is (a)/(b)/(c)/(d) Solution: (d) Assertion(A) is false and Reasoning(R) is true.Assertion: In this situation given in the diagram, the sum of opposite sides is always equal.So, the correct relation should be: AB + CD = AD + CBHence, the assertion is false.Reasoning: We know that in two concentric circles, the chord of the larger circle, which touches( or acts as a tangent to) the smaller circle, is bisected at the point of contact.Therefore, Reasoning (R) is correct....

Two cars are approaching each other at an equal speed of 7.2 km / hr.

Question: Two cars are approaching each other at an equal speed of $7.2 \mathrm{~km} / \mathrm{hr}$. When they see each other, both blow horns having frequency of $676 \mathrm{Hz.}$. The beat frequency heard by each driver will be_____ Hz. [Velocity of sound in air is $340 \mathrm{~m} / \mathrm{s}$.] Solution: Speed $=7.2 \mathrm{~km} / \mathrm{h}=2 \mathrm{~m} / \mathrm{s}$ Frequency as heard by A $f_{A}^{\prime}=f_{B}\left(\frac{V+V_{0}}{V-V_{n}}\right)$ $f_{A}^{\prime}=676\left(\frac{340+2}{3...

Assertion (A) If two tangents are drawn to a circle from an external point,

Question: Assertion (A)If two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.Reason (R)A parallelogram circumscribing a circle is a rhombus.(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).(c) Assertion (A) is true and Reason (R) is false.(d) Assertion (A) is false and Reason (R)is true. ...

The value of [(sec A + tan A) (1−sin A)] is equal to

Question: The value of [(sec A + tan A) (1sin A)] is equal to (a) tan2A(b) sin2A(c) cos A(d) sin A Solution: Here we have to evaluate the value of $(\sec A+\tan A)(1-\sin A)$ Now we are going to solve this $(\sec A+\tan A)(1-\sin A)$ $=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$ (here we are using the trigonometrical relation) $=\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)$ $=\frac{1}{\cos A}(1+\sin A)(1-\sin A)$ $=\frac{1}{\cos A}\left(1-\sin ^{2} A\right)$ we know that $(a...

Write the additive inverse of each of the following rational numbers:

Question: Write the additive inverse of each of the following rational numbers: (i) $\frac{-2}{17}$ (ii) $\frac{3}{-11}$ (iii) $\frac{-17}{5}$ (iv) $\frac{-11}{-25}$ Solution: (i) Additive inverse is the negative of the given number. So, additive inverse of $\frac{-2}{17}$ is $\frac{2}{17}$. (ii) Additive inverse is the negative of the given number. So, additive inverse of $\frac{3}{-11}$ is $\frac{3}{11}$. (iii) Additive inverse is the negative of the given number. So, additive inverse of $\fra...

Which of the following equations represents a travelling wave?

Question: Which of the following equations represents a travelling wave?$y=A e^{-x^{2}}(v t+\theta)$$y=A \sin (15 x-2 t)$$y=A e^{x} \cos (\omega t-\theta)$$y=A \sin x \cos \omega t$Correct Option: , 2 Solution: (2) $Y=F(x, t)$ For travelling wave $y$ should be linear function of $x$ and $t$ and they must exist as $(x \pm v t) Y=A \sin (15 x-2 t) \rightarrow$ linear function in $x$ and $t$...

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