The base of an isosceles triangle is 16 cm and its area is 48 cm2.

Question: The base of an isosceles triangle is 16 cm and its area is 48 cm2. The perimeter of the triangle is(a) 41 cm(b) 36 cm(c) 48 cm(d)324 cm Solution: (b) 36 cm Let $\triangle P Q R$ be an isosceles triangle and $P X \perp Q R$. Now, Area of triangle $=48 \mathrm{~cm}^{2}$ $\Rightarrow \frac{1}{2} \times Q R \times P X=48$ $\Rightarrow h=\frac{96}{16}=6 \mathrm{~cm}$ Also, $Q X=\frac{1}{2} \times 24=12 \mathrm{~cm}$ and $P X=12 \mathrm{~cm}$ $P Q=\sqrt{Q X^{2}+P X^{2}}$ $a=\sqrt{8^{2}+6^{2}...

Let f(x) be a polynomial of degree 3 such that

Question: Let $f(x)$ be a polynomial of degree 3 such that $f(-1)=10$, $f(1)=-6, f(x)$ has a critical point at $x=-1$ and $f^{\prime}(x)$ has a critical point at $x=1$. Then $f(x)$ has a local minima at $x=$ Solution: Let $f(x)=a x^{3}+b x^{2}+c x+d$ $f(-1)=10$ and $f(1)=-6$ $-a+b-c+d=10$........(1) $a+b+c+d=-6$........(2) Solving equations (i) and (ii), we get $a=\frac{1}{4}, d=\frac{35}{4}$ $b=\frac{-3}{4}, c=-\frac{9}{4}$ $\Rightarrow f(x)=a\left(x^{3}-3 x^{2}-9 x\right)+d$ $f^{\prime}(x)=\fr...

An AC circuit has

Question: An AC circuit has $R=100 \Omega, C=2 \mu \mathrm{F}$ and $L=80 \mathrm{mH}$, connected in series. The quality factor of the circuit is :(1) 2(2) $0.5$(3) 20(4) 400Correct Option: 1 Solution: (1) Quality factor, $Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{100} \sqrt{\frac{80 \times 10^{-3}}{2 \times 10^{-6}}}$ $=\frac{1}{100} \sqrt{40 \times 10^{3}}=\frac{200}{100}=2$...

Assertion A: Enol form of acetone

Question: Assertion A: Enol form of acetone $\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]$ exists in $0.1 \%$ quantity. However, the enol form of acetyl acetone $\left[\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{OCCH}_{3}\right]$ exists in approximately $15 \%$ quantity. Reason $\mathrm{R}$ : enol form of acetyl acetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone. Choose the correct statement:$A$ is false but $R$ is trueBoth $A$ and $R$ are tru...

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively.

Question: The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is(a) 24 cm(b) 18 cm(c) 30 cm(d)12 cm Solution: (a) 24 cm Let: $a=30 \mathrm{~cm}, b=24 \mathrm{~cm}$ and $c=18 \mathrm{~cm}$ $s=\frac{a+b+c}{2}=\frac{30+24+18}{2}=36 \mathrm{~cm}$ On applying Heron's formula, we get: Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{36(36-30)(36-24)(36-18)}$ $=\sqrt{36 \times 6 \times 12 ...

Question: Assertion A: Enol form of acetone $\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]$ exists in $0.1 \%$ quantity. However, the enol form of acetyl acetone $\left[\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{OCCH}_{3}\right]$ exists in approximately $15 \%$ quantity. Reason $\mathrm{R}$ : enol form of acetyl acetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone. Choose the correct statement:$A$ is false but $R$ is trueBoth $A$ and $R$ are tru...

A series L-R circuit is connected to a battery of emf V.

Question: A series $L-R$ circuit is connected to a battery of emf $V$. If the circuit is switched on at $t=0$, then the time at which the energy stored in the inductor reaches $\left(\frac{1}{n}\right)$ times of its maximum value, is:(1) $\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$(2) $\frac{L}{R} \ln \left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)$(3) $\frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)$(4) $\frac{L}{R} \ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$Correct Option: ...

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm.

Question: The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is(a) 375 cm2(b) 750 cm2(c) 250 cm2(d)500 cm2 Solution: (b) $750 \mathrm{~cm}^{2}$ Let the sides of the triangle be 5xcm, 12xcm and 13xcm.Perimeter = Sum of all sidesor, 150 = 5x+ 12x+ 13xor, 30x= 150or,x= 5Thus, the sides of the triangle are 55 cm, 125 cm and 135 cm, i.e., 25 cm, 60 cm and 65 cm.Now Let : $a=25 \mathrm{~cm}, b=60 \mathrm{~cm}$ and $c=65 \mathrm{~cm}$ $s=\frac{150...

Let be the set of all functions

Question: Let $\mathrm{S}$ be the set of all functions $f:[0,1] \rightarrow R$, which are continuous on $[0,1]$ and differentiable on $(0,1)$. Then for every $f$ in $S$, there exists a $c \in(0,1)$, depending on $f$, such that:(1) $|f(c)-f(1)|(1-c)\left|f^{\prime}(c)\right|$(2) $\frac{f(1)-f(c)}{1-c}=f^{\prime}(c)$(3) $|f(c)+f(1)|(1+c)\left|f^{\prime}(c)\right|$(4) None of theseCorrect Option: , 4 Solution: For a constant function $f(x)$, option (1), (3) doesn't hold and by LMVT theorem, option ...

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is

Question: The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is(a) 480 m2(b) 320 m2(c) 384 m2(d) 360 m2 Solution: (c) $384 \mathrm{~m}^{2}$ Let: $a=40 \mathrm{~m}, b=24 \mathrm{~m}$ and $c=32 \mathrm{~m}$ $s=\frac{a+b+c}{2}=\frac{40+24+32}{2}=48 \mathrm{~m}$ By Heron's formula, we have : Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{48(48-40)(48-24)(48-32)}$ $=\sqrt{48 \times 8 \times 24 \times 16}$ $=\sqrt{24 \times 2 \times...

The length of the perpendicular from the origin,

Question: The length of the perpendicular from the origin, on the normal to the curve, $x^{2}+2 x y-3 y^{2}=0$ at the point $(2,2)$ is:(1) $\sqrt{2}$(2) $4 \sqrt{2}$(3) 2(4) $2 \sqrt{2}$Correct Option: , 4 Solution: Given equation of curve is $x^{2}+2 x y-3 y^{2}=0$ $\Rightarrow 2 x+2 y+2 x y^{\prime}-6 y y^{\prime}=0$ $\Rightarrow x+y+x y^{\prime}-3 y y^{\prime}=0$ $\Rightarrow y^{\prime}(x-3 y)=-(x+y)$ $\Rightarrow \frac{d y}{d x}=\frac{x+y}{3 y-x}$ Slope of normal $=\frac{-d x}{d y}=\frac{x-3...

A galvanometer coil has 500 turns and each turn has an average area

Question: A galvanometer coil has 500 turns and each turn has an average area of $3 \times 10^{-4} \mathrm{~m}^{2}$. If a torque of $1.5 \mathrm{Nm}$ is required to keep this coil parallel to a magnetic field when a current of $0.5 \mathrm{~A}$ is flowing through it, the strength of the field (in $\mathrm{T}$ ) is Solution: (20) Given, Area of galvanometer coil, $A=3 \times 10^{-4} \mathrm{~m}^{2}$ Number of turns in the coil, $N=500$ Current in the coil, $I=0.5 \mathrm{~A}$ Torque $\tau=|\vec{M...

The height of an equilateral triangle is 6 cm. Its area is

Question: The height of an equilateral triangle is 6 cm. Its area is (a) $12 \sqrt{3} \mathrm{~cm}^{2}$ (b) $6 \sqrt{3} \mathrm{~cm}^{2}$ (c) $12 \sqrt{2} \mathrm{~cm}^{2}$ (d) $18 \mathrm{~cm}^{2}$ Solution: (a) $12 \sqrt{3} \mathrm{~cm}^{2}$ Height of equilateral triangle $=\frac{\sqrt{3}}{2} \times$ Side $\Rightarrow 6=\frac{\sqrt{3}}{2} \times$ Side $\Rightarrow$ Side $=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{12}{3} \times \sqrt{3}=4 \sqrt{3} \mathrm{~cm}$ Now, Area of equ...

For a>0, let the curves

Question: For $a0$, let the curves $C_{1}: y^{2}=a x$ and $C_{2}: x^{2}=a y$ intersect at origin $O$ and a point $P$. Let the line $x=b(0ba)$ intersect the chord $O P$ and the $x$-axis at points $Q$ and $R$, respectively. If the line $x=b$ bisects the area bounded by the curves, $C_{1}$ and $C_{2}$, and the area of $\Delta O Q R=\frac{1}{2}$, then ' $a$ ' satisfies the equation:(1) $x^{6}-6 x^{3}+4=0$(2) $x^{6}-12 x^{3}+4=0$(3) $x^{6}+6 x^{3}-4=0$(4) $x^{6}-12 x^{3}-4=0$Correct Option: , 2 Solut...

A 750 Hz}, 20V rms source is connected to a resistance of 100

Question: A $750 \mathrm{~Hz}, 20 \mathrm{~V}$ (rms) source is connected to a resistance of $100 \Omega$, an inductance of $0.1803 \mathrm{H}$ and a capacitance of $10 \mu \mathrm{F}$ all in series. The time in which the resistance (heat capacity $2 \mathrm{~J} /{ }^{\circ} \mathrm{C}$ ) will get heated by $10^{\circ} \mathrm{C}$. (assume no loss of heat to the surroundings) is close to :(1) $418 \mathrm{~s}$(2) $245 \mathrm{~s}$(3) $365 \mathrm{~s}$(4) $348 \mathrm{~s}$Correct Option: , 4 Solut...

Each side of an equilateral triangle is 10 cm long.

Question: Each side of an equilateral triangle is 10 cm long. The height of the triangle is (a) $10 \sqrt{3} \mathrm{~cm}$ (b) $5 \sqrt{3} \mathrm{~cm}$ (c) $10 \sqrt{2} \mathrm{~cm}$ (d) $5 \mathrm{~cm}$ Solution: (b) $5 \sqrt{3} \mathrm{~cm}$ Height of equilateral triangle $=\frac{\sqrt{3}}{2} \times$ Side $=\frac{\sqrt{3}}{2} \times 10$ $=5 \sqrt{3} \mathrm{~cm}$...

In the following reaction

Question: In the following reaction $\mathrm{CH}_{3} \mathrm{CHO}$ and ${ }^{\mathrm{t}} \mathrm{BuOH}$$\mathrm{HCHO}$ and $\mathrm{MeOH}$$\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{MeOH}$$\mathrm{HCHO}$ and ${ }^{\mathrm{t}} \mathrm{BuOH}$Correct Option: , 2 Solution: Best combination is HCHO (more reactive aldehyde) and $\mathrm{MeOH}$ (less sterically hindered alcohol)....

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is

Question: Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is (a) $5 \sqrt{10} \mathrm{~cm}^{2}$ (b) $50 \mathrm{~cm}^{2}$ (c) $10 \sqrt{3} \mathrm{~cm}^{2}$ (d) $75 \mathrm{~cm}^{2}$ Solution: (b) 50 cm2Here, the base and height of the triangle are 10 cm and 10 cm, respectively.Thus, we have: Area of triangle $=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times 10 \times 10$ $=50 \mathrm{~cm}^{2}$...

then which of the following is true?

Question: Let $f(x)=x \cos ^{-1}(-\sin |x|), x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then which of the following is true?(1) $f^{\prime}$ is increasing in $\left(-\frac{\pi}{2}, 0\right)$ and decreasing in $\left(0, \frac{\pi}{2}\right)$(2) $f^{\prime}(0)=-\frac{\pi}{2}$(3) $f^{\prime}$ is not differentiable at $x=0$(4) $f^{\prime}$ is decreasing in $\left(-\frac{\pi}{2}, 0\right)$ and increasing in $\left(0, \frac{\pi}{2}\right)$Correct Option: , 4 Solution: $f^{\prime}(x)=x\left(\pi-...

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm.

Question: The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is (a) 8 cm (b) $\sqrt{30} \mathrm{~cm}$ (c) $4 \mathrm{~cm}$ (d) $\sqrt{11} \mathrm{~cm}$ Solution: (c) 4 cm Height of isosceles triangle $=\frac{1}{2} \sqrt{4 a^{2}-b^{2}}$ $=\frac{1}{2} \sqrt{4(5)^{2}-6^{2}} \quad(a=5 \mathrm{~cm}$ and $b=6 \mathrm{~cm})$ $=\frac{1}{2} \times \sqrt{100-36}$ $=\frac{1}{2} \times \sqrt{64}$ $=\frac{1}{2} \times 8$ $=4 \mathrm{~cm}$...

The major product of the following reaction is :

Question: The major product of the following reaction is : Correct Option: , 4 Solution:...

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm.

Question: The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is (a) $16 \sqrt{5} \mathrm{~cm}^{2}$ (b) $8 \sqrt{5} \mathrm{~cm}^{2}$ (c) $16 \sqrt{3} \mathrm{~cm}^{2}$ (d) $8 \sqrt{3} \mathrm{~cm}^{2}$ Solution: (b) $8 \sqrt{5} \mathrm{~cm}^{2}$ Area of isosceles triangle $=\frac{b}{4} \sqrt{4 a^{2}-b^{2}}$ Here, $a=6 \mathrm{~cm}$ and $b=8 \mathrm{~cm}$ Thus, we have: $\frac{8}{4} \times \sqrt{4(6)^{2}-8^{2}}$ $=\frac{8}{4} \times ...

If c is a point at which Rolle's theorem holds for the function

Question: If $c$ is a point at which Rolle's theorem holds for the function, $f(x)=\log _{e}\left(\frac{x^{2}+a}{7 x}\right)$ in the interval $[3,4]$, where $\alpha \in R$, then $f^{\prime \prime}(c)$ is equal to:(1) $-\frac{1}{12}$(2) $\frac{1}{12}$(3) $-\frac{1}{24}$(4) $\frac{\sqrt{3}}{7}$Correct Option: , 2 Solution: Since, Rolle's theorem is applicable $\therefore f(a)=f(b)$ $f(3)=f(4) \Rightarrow \alpha=12$ $f^{\prime}(x)=\frac{x^{2}-12}{x\left(x^{2}+12\right)}$ As $f^{\prime}(c)=0$ (by Ro...

The major product of the following reaction is :

Question: The major product of the following reaction is : Correct Option: 1 Solution:...

Each side of an equilateral triangle measure 8 cm.

Question: Each side of an equilateral triangle measure 8 cm. The area of the triangle is (a) $8 \sqrt{3} \mathrm{~cm}^{2}$ (b) $16 \sqrt{3} \mathrm{~cm}^{2}$ (c) $32 \sqrt{3} \mathrm{~cm}^{2}$ (d) $48 \mathrm{~cm}^{2}$ Solution: (b) $16 \sqrt{3} \mathrm{~cm}^{2}$ Area of equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$ $=\frac{\sqrt{3}}{4} \times(8)^{2}$ $=\frac{\sqrt{3}}{4} \times 64$ $=16 \sqrt{3} \mathrm{~cm}^{2}$...