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

If sin 3θ = cos (θ − 6°),

Question: If sin 3 = cos ( 6), where (3) and ( 6) are both acute angles, then the value of is(a) 18(b) 24(c) 36(d) 30 Solution: Given that, $\sin 3 \theta=\cos (\theta-6)$ (1) We have to find Here the angles (3) and ( 6) are acute angles and we know that Therefore we can rewrite the equation (1) as $\cos (90-3 \theta)=\cos (\theta-6)$ $\Rightarrow \quad 90-3 \theta=\theta-6$ $\Rightarrow \quad 4 \theta=96$ $\Rightarrow \quad \theta=\frac{96}{4}=24^{\circ}$ Therefore option $(b)$ is correct....

Simplify:

Question: Simplify: (i) $\left(\frac{25}{8} \times \frac{2}{5}\right)-\left(\frac{3}{5} \times \frac{-10}{9}\right)$ (ii) $\left(\frac{1}{2} \times \frac{1}{4}\right)+\left(\frac{1}{2} \times 6\right)$ (iii) $\left(-5 \times \frac{2}{15}\right)-\left(-6 \times \frac{2}{9}\right)$ (iv) $\left(\frac{-9}{4} \times \frac{5}{3}\right)+\left(\frac{13}{2} \times \frac{5}{6}\right)$ (v) $\left(\frac{-4}{3} \times \frac{12}{-5}\right)+\left(\frac{3}{7} \times \frac{21}{15}\right)$ (vi) $\left(\frac{13}{5...

Draw two tangents to a circle of radius 3.5 cm from a point P

Question: Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre. Solution: Steps of ConstructionStep 1. Draw a circle with O as centre and radius 3.5 cm.Step 2. Mark a point P outside the circle such that OP = 6.2 cm.Step 3. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.Step 4. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.Step 5. Join PT and PT'. Here, PT and PT' are...

A cricket ball of mass 0.15 kg is thrown vertically

Question: A cricket ball of mass $0.15 \mathrm{~kg}$ is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 $\mathrm{m}$ after leaving the machine. If the part pushing the ball applies a constant force $F$ on the ball and moves horizontally a distance of $0.2 \mathrm{~m}$ while launching the ball, the value of $F$ (in N) is $\left(g=10 \mathrm{~ms}^{-2}\right)$______ Solution: $(150.00)$ From work energy theorem, $W=F \cdot s=\Delta K E=\frac{1}{2} m v^{2}$ Here ...

In Fig. 2, If DE || BC, then x equals

Question: In Fig. 2, If DE || BC, then x equals(a) 6 cm(b) 8 cm(c) 10 cm(d) 12.5 cm Solution: In the given figure we have DE||BC. We have to findx. In the following given figure DE||BC, so triangle ADE and triangle ABC are similar triangle. So we have the following relation $\frac{B C}{D E}=\frac{A B}{A D}$ $\frac{x}{4}=\frac{3+2}{2}$ $\Rightarrow x=10$ Hence the correct option is...

Draw a circle of radius 3 cm. From a point P, 7 cm away from the centre of the circle,

Question: Draw a circle of radius 3 cm. From a point P, 7 cm away from the centre of the circle, draw two tangents to the circle. Also, measure the lengths of the tangents. Solution: Steps of ConstructionStep 1. Draw a circle with O as centre and radius 3 cm.Step 2. Mark a point P outside the circle such that OP = 7 cm.Step 3. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.Step 4. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the point...

Draw a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm.

Question: Draw a right triangle in which sides (other than hypotenuse) are of lengths $4 \mathrm{~cm}$ and $3 \mathrm{~cm}$. Then, construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle. Solution: Steps of ConstructionStep 1. Draw a line segment BC = 3 cm.Step 2. At B, drawXBC = 90.Step 3. With B as centre and radius 4 cm, draw an arc cutting BX at A.Step 4. Join AC. Thus, a right ∆ABC is obtained. Step 5. Extend BC to D such that BD $=\frac{...

A small block starts slipping down from a point B on an inclined plane A B,

Question: A small block starts slipping down from a point $B$ on an inclined plane $A B$, which is making an angle $\theta$ with the horizontal section $B C$ is smooth and the remaining section $C A$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$ ) of the inclined plane. If $B C=2 A C$, the coefficient of friction is given by $\mu=k \tan \theta$. The value of $k$ is______ Solution: (3) If $A C=l$ then according to quest...

Simplify each of the following and express the result as a rational number in standard form:

Question: Simplify each of the following and express the result as a rational number in standard form: (i) $\frac{-16}{21} \times \frac{14}{5}$ (ii) $\frac{7}{6} \times \frac{-3}{28}$ (iii) $\frac{-19}{36} \times 16$ (iv) $\frac{-13}{9} \times \frac{27}{-26}$ (v) $\frac{-9}{16} \times \frac{-64}{-27}$ (vi) $\frac{-50}{7} \times \frac{14}{3}$ (vii) $\frac{-11}{9} \times \frac{-81}{-88}$ (viii) $\frac{-5}{9} \times \frac{72}{-25}$ Solution: (i) $\frac{-16}{21} \times \frac{14}{5}=\frac{-2 \times 2...

Simplify each of the following and express the result as a rational number in standard form:

Question: Simplify each of the following and express the result as a rational number in standard form: (i) $\frac{-16}{21} \times \frac{14}{5}$ (ii) $\frac{7}{6} \times \frac{-3}{28}$ (iii) $\frac{-19}{36} \times 16$ (iv) $\frac{-13}{9} \times \frac{27}{-26}$ (v) $\frac{-9}{16} \times \frac{-64}{-27}$ (vi) $\frac{-50}{7} \times \frac{14}{3}$ (vii) $\frac{-11}{9} \times \frac{-81}{-88}$ (viii) $\frac{-5}{9} \times \frac{72}{-25}$ Solution: (i) $\frac{-16}{21} \times \frac{14}{5}=\frac{-2 \times 2...

Two masses A and B, each of mass M are fixed together by a massless springs.

Question: Two masses $A$ and $B$, each of mass $M$ are fixed together by a massless springs. A force acts on the mass $\mathrm{B}$ as shown in figure. If the mass $\mathrm{A}$ starts moving away from mass $\mathrm{B}$ with acceleration ' $\mathrm{a}$ ', than the acceleration of mass $\mathrm{B}$ will be: $\frac{\mathrm{F}+\mathrm{Ma}}{\mathrm{M}}$$\frac{\mathrm{F}-\mathrm{Ma}}{\mathrm{M}}$$\frac{M a-F}{M}$$\frac{M F}{F+M a}$Correct Option: , 2 Solution: $F-F_{s}=M a^{\prime}$ $a^{\prime}=\frac{F...

Multiply:

Question: Multiply: (i) $\frac{-5}{17}$ by $\frac{51}{-60}$ (ii) $\frac{-6}{11}$ by $\frac{-55}{36}$ (iii) $\frac{-8}{25}$ by $\frac{-5}{16}$ (iv) $\frac{6}{7}$ by $\frac{-49}{36}$ (v) $\frac{8}{-9}$ by $\frac{-7}{-16}$ (vi) $\frac{-8}{9}$ by $\frac{3}{64}$ Solution: (i) $\frac{-5}{17} \times \frac{51}{-60}=\frac{-5}{17} \times \frac{3 \times 17}{-5 \times 3 \times 4}=\frac{-5 \times 3 \times 17}{-17 \times 5 \times 3 \times 4}=\frac{1}{4}$ (ii)$\frac{-6}{11} \times \frac{-55}{36}=\frac{-6}{11} ...

Two particles having masses 4g and 16g respectively are moving with equal kinetic energies.

Question: Two particles having masses $4 \mathrm{~g}$ and $16 \mathrm{~g}$ respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is $\mathrm{n}: 2$. The value of $\mathrm{n}$ will be Solution: (1) $\because$ relation $\mathrm{b} / \mathrm{w}$ kinetic energy $\backslash$ momentum is $\mathrm{P}=\sqrt{2 \mathrm{mKE}} \quad(\because \mathrm{KE}=$ same $)$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}$ $\frac{...

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance)

Question: The potential energy (U) of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $\mathbf{U}=\frac{\alpha}{r^{10}}-\frac{\beta}{r^{5}}-3$ Where, $\mathbf{a}$ and $\mathbf{b}$ are positive constants. The equilibrium distance between two atoms will $\left(\frac{2 \alpha}{\beta}\right)^{\frac{a}{b}} \cdot$ Where $\mathbf{a}=$ Solution: (1) $F=-\frac{d U}{d r}$ $F=-\left[-\frac{10 \alpha}{r^{11}}+\frac{5 \beta}{r^{6}}\right]$ for equilibrium, $\mathrm{F}=0$ $\frac{1...

Multiply:

Question: Multiply: (i) $\frac{7}{11}$ by $\frac{5}{4}$ (ii) $\frac{5}{7}$ by $\frac{-3}{4}$ (iii) $\frac{-2}{9}$ by $\frac{5}{11}$ (iv) $\frac{-3}{17}$ by $\frac{-5}{-4}$ (v) $\frac{9}{-7}$ by $\frac{36}{-11}$ (vi) $\frac{-11}{13}$ by $\frac{-21}{7}$ (vii) $-\frac{3}{5}$ by $-\frac{4}{7}$ (viii) $-\frac{15}{11}$ by 7 Solution: (i)$\frac{7}{11} \times \frac{5}{4}=\frac{7 \times 5}{11 \times 4}=\frac{35}{44}$ (ii) $\frac{5}{7} \times \frac{-3}{4}=\frac{5 \times(-3)}{7 \times 4}=\frac{-15}{28}$ (i...

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm

Question: Construct an isosceles triangle whose base is $8 \mathrm{~cm}$ and altitude $4 \mathrm{~cm}$ and then another triangle whose sides are $1 \frac{1}{2}$ times the corresponding sides of the isosceles triangle. Solution: Steps of ConstructionStep 1. Draw a line segment BC = 8 cm.Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D.Step 3. With D as centre and radius 4 cm, draw an arc cutting XY at A.Step 4. Join AB and AC. Thus, an isosceles ∆ABC whose base is 8 cm and altitu...

To construct a triangle similar to ∆ABC in which BC = 4.5 cm,

Question: To construct a triangle similar to $\triangle \mathrm{ABC}$ in which $\mathrm{BC}=4.5 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ}$ and $\angle \mathrm{C}=60^{\circ}$, using a scale factor of $\frac{3}{7}, \mathrm{BC}$ will be divided in the ratio (a) $3: 4$ (b) $4: 7$ (c) $3: 10$ (d) $3: 7$ Solution: To construct a triangle similar to $\triangle \mathrm{ABC}$ in which $\mathrm{BC}=4.5 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ}$ and $\angle \mathrm{C}=60^{\circ}$, using a scale factor of...

Two solids A and B of mass 1kg and 2kg

Question: Two solids $\mathrm{A}$ and $\mathrm{B}$ of mass $1 \mathrm{~kg}$ and $2 \mathrm{~kg}$ respectively are moving withequal linear momentum. The ratio of their kinetic energies $(\mathrm{K} . \mathrm{E} .)_{\mathrm{A}}:(\mathrm{K} . \mathrm{E} .)_{\mathrm{B}}$ will be $\frac{\mathrm{A}}{1} .$ So the value of $\mathrm{A}$ will be Solution: (2) Given that, $\frac{M_{1}}{M_{2}}=\frac{1}{2}$ Also, $\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{p}$ $\Rightarrow \mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M...

Simplify:

Question: Simplify: (i) $\frac{-3}{2}+\frac{5}{4}-\frac{7}{4}$ (ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}{3}$ (iii) $\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$ (iv) $\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7}$ (v) $\frac{5}{6}+\frac{-2}{5}-\frac{-2}{15}$ (vi) $\frac{3}{8}-\frac{-2}{9}+\frac{-5}{36}$ Solution: (i) \frac{-3}{2}+\frac{5}{4}-\frac{7}{4} Taking the L.C.M. of the denominators: $\frac{-6}{4}+\frac{5}{4}-\frac{7}{4}$ $=\frac{-6+5-7}{4}$ $=\frac{-8}{4}$ $=-2$ (ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}...

Construct a ∆ABC in which BC = 8 cm, ∠B = 45º and ∠C = 60º.

Question: Construct a $\triangle \mathrm{ABC}$ in which $\mathrm{BC}=8 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ}$ and $\angle \mathrm{C}=60^{\circ} .$ Construct another triangle similar to $\triangle \mathrm{ABC}$ such that its sides are $\frac{3}{5}$ of the corresponding sides of ∆ABC. Solution: Steps of ConstructionStep 1. Draw a line segment BC = 8 cm.Step 2. At B, drawXBC = 45.Step 3. At C, drawYCB = 60. Suppose BX and CY intersect at A.Thus, ∆ABC is the required triangle.Step 4. Below BC, ...

A ball of mass 4 kg,

Question: A ball of mass $4 \mathrm{~kg}$, moving with a velocity of $10 \mathrm{~ms}^{-1}$, collides with a spring of length $8 \mathrm{~m}$ and force constant $100 \mathrm{Nm}^{-1}$. The length of the compressed spring is $x \mathrm{~m}$. The value of $\mathrm{x}$, to the nearest integer, is Solution: (6) Let's say the compression in the spring by: $y$. So, by work energy theorem we have $\Rightarrow \frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{ky}^{2}$ $\Rightarrow \mathrm{y}=\sqrt{\frac{\...

Construct a ∆ABC in which AB = 6 cm, ∠A = 30º and ∠B = 60º.

Question: Construct a $\triangle \mathrm{ABC}$ in which $\mathrm{AB}=6 \mathrm{~cm}, \angle \mathrm{A}=30^{\circ}$ and $\angle \mathrm{B}=60^{\circ} .$ Construct another $\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime}$ similar to $\Delta \mathrm{ABC}$ with base $\mathrm{AB}^{\prime}=8 \mathrm{~cm}$. Solution: Steps of ConstructionStep 1. Draw a line segment AB = 6 cm.Step 2. At A, drawXAB = 30.Step 3. At B, drawYBA = 60. Suppose AX and BY intersect at C.Thus, ∆ABC is the required triangle.Step ...

Simplify:

Question: Simplify: (i) $\frac{-3}{2}+\frac{5}{4}-\frac{7}{4}$ (ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}{3}$ (iii) $\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$ (iv) $\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7}$ (v) $\frac{5}{6}+\frac{-2}{5}-\frac{-2}{15}$ (vi) $\frac{3}{8}-\frac{-2}{9}+\frac{-5}{36}$ Solution: (i) \frac{-3}{2}+\frac{5}{4}-\frac{7}{4} Taking the L.C.M. of the denominators: $\frac{-6}{4}+\frac{5}{4}-\frac{7}{4}$ $=\frac{-6+5-7}{4}$ $=\frac{-8}{4}$ $=-2$ (ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}...

Construct a ΔABC with BC = 6 cm, ∠B = 60° and AB = 5 cm.

Question: Construct a $\triangle A B C$ with $B C=6 \mathrm{~cm}, \angle B=60^{\circ}$ and $A B=5 \mathrm{~cm}$. Construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of $\triangle A B C$. Solution: Given: InΔABC,BC= 6 cm,B = 60AB= 5 cmSteps of construction:(1) Draw a line segmentAB= 5 cm.(2) From the pointB, draw an ABY=60(3) TakingBas center, 6 cm radius, draw an arc on the rayBY, Let the point where the arc intersects the ray named asC.(4) JoinAC. Hence,ΔABC i...

A particle of mass $m$ moves in a circular orbit under the central potential

Question: A particle of mass $m$ moves in a circular orbit under the central potential field, $U(r)=\frac{-C}{r}$, where $C$ is a positive constant. The correct radius - velocity graph of the particle's motion is :Correct Option: 1 Solution: (1) $\mathrm{U}=-\frac{\mathrm{C}}{\mathrm{r}}$ $\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dr}}=-\frac{\mathrm{C}}{\mathrm{r}^{2}}$ $|\mathrm{F}|=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\frac{\mathrm{C}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\mathrm...

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