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Class 12th Quantitative Aptitude Prep Tips for MBA

f(x) is a quadratic polynomial with maximum value of 3 at x = 2. If f (0) = 2. Find f (6)

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Payal Gupta

Contributor-Level 10

Let f (x) = ax² + bx + c

f (0)= 2

2 = a (0)² + b (0) + c

c = 2

f (– 2) = 3

a (–2)² +b (–2) + 2 = 3

4a – 2b = 1 ………. (1)

The given equation has a maximum value at x = –2

$\frac{- b}{2 a} = - 2$

b = 4a

put in equation 1

we get, $a = \frac{- 1}{4}, b = - 1$

f (6)= $\frac{- 1}{4} {(6)}^{2} + (- 1) 6 + 2$

= –13

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7 months ago

Class 12th Quantitative Aptitude Prep Tips for MBA

ABCD is a square, having side 12cm, in which ABC and ABD are two quadrant of a circle. If a circle is drawn touching both the quadrants and the square as shown in figure as show in figure given below, then find the area of circle.

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Payal Gupta

Contributor-Level 10

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7 months ago

Class 12th Physics Wave Optics

An unpolarised light beam strikes a glass surface at Brewster's angle. Then

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alok kumar singh

Contributor-Level 10

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Class 12th Quantitative Aptitude Prep Tips for MBA

If P₁ + P₂ + P₃ = 59, where P₁, P₂ and P₃ are prime numbers and P₁ < P₂ < P₃, then how many values can P₃ take?

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Payal Gupta

Contributor-Level 10

P₁ + P₂ + P₃ = 59

P₁ and P₂ → minimize

Then P₃ → minimum i.e. 47

P₁	P₂	P₃
5	7	47
5	11	43
3	13	43
5	13	41
5	17	37
3	19	37
5	23	31
7	23	29

So, only possible value of P₃ can take → 29, 31, 37, 41, 43, and 47

Total 6 values

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7 months ago

Class 12th Physics Magnetism and Matter

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 * 1 0^{- 6} kg m^{2}$ . If the magnitude of magnetic moment of the needle is $x * 1 0^{- 5} A m^{2}$ , then the value of ' $x$ ' is :

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alok kumar singh

Contributor-Level 10

Time period of Oscillation, $T = 2 π \sqrt[]{\frac{I}{M B}}$

$\Rightarrow \frac{1}{4} = 2 π \sqrt[]{\frac{9.8 * 1 0^{- 6}}{M * 0.049}}$

$\Rightarrow \frac{1}{16} = 4 π^{2} * \frac{9.8 * 1 0^{- 6}}{M * 49 * 1 0^{- 3}}$

$\Rightarrow M = \frac{4 π^{2} * 9.8 * 1 0^{- 6}}{49 * 1 0^{- 3}} * 16$

$= \frac{4 π^{2} * 9.8 * 16 * 1 0^{- 3}}{49}$

$= 12.8 π^{2} * 1 0^{- 3} * 1 0^{- 2} * 1 0^{2}$

$= 1280 π^{2} * 1 0^{- 5} A m^{2}$

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7 months ago

Class 12th Quantitative Aptitude Prep Tips for MBA

How many pairs of positive integers x, y satisfy $\frac{1}{x} + \frac{6}{y} = \frac{1}{15}$ , where y is a two digit odd integer?

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Payal Gupta

Contributor-Level 10

$\frac{1}{x} = \frac{1}{15} - \frac{6}{y}$

$\frac{1}{x} = \frac{y - 90}{15 y}$

$x = \frac{15 y}{y - 90}$ ______equation 1

i.e. y = 90+y,

1≤ y₁ ≤9

y₁ is odd

Equation 1 becomes

$x = \frac{15 (90 + y_{1})}{y_{1}}$

$= 15 + \frac{90 * 15}{y_{1}}$

y₁ can take 1,3,5,9

y=91, 93, 95, 99

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7 months ago

Class 12th Quantitative Aptitude Prep Tips for MBA

An alcohol solution is formed by mixing two solutions. Solution 1 had a concentration of 20 percent and a volume of 3L while solution 2 had a concentration of 40 percent with a volume of 1L. The resultant solution was named solution 3. Three litres of solution 4 which had an unknown concentration was mixed with solution 3. What must be concentration of solution 4 if that the new resultant solution has a concentration of 50 percent?

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Payal Gupta

Contributor-Level 10

Sol1	20percent (0.6)	3L
Sol2	40percent (0.4)	1L
Sol3	1L	4L
Sol4	XL	3L
Sol5	3.5L	7L

1 + x = 3.5

x= 2.5

Required percent y = $\frac{2.5}{3} * 100$ = 83.33 percent

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7 months ago

Class 12th Quantitative Aptitude Prep Tips for MBA

Two geometric means a, b are inserted between x and y such that a³+ b³ = n (ab) A,where A is the arithmetic mean of x and y. What is value of n?

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Payal Gupta

Contributor-Level 10

x, a, b, y are in GP

a = xr

b = xr²

y = xr³

r = ${(\frac{y}{x})}^{1 / 3}$

rewrite,

a = x^2/3.y^1/3

b = x^1/3.y^2/3

a * b = x * y

a³+ b³= x²y + y² x

= xy (x+y)

= xy (2A) …………… $[\frac{x + y}{2} = A]$

Hence, n = 2

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7 months ago

Class 12th Quantitative Aptitude Prep Tips for MBA

A square with maximum possible length of the side is inscribed in an equilateral triangle and a circle of maximum possible radius is inscribed in the square. Find the ratio of the length of equilateral of the equilateral triangle to the radius of circle.

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Payal Gupta

Contributor-Level 10

Let the side of = a

CD = a/2

Let EC = x

In GEC

GE/EC = tan60 = $\sqrt[]{3}$

GE = x $\sqrt[]{3}$

DE = a/2 – x = FG = HF

GE = HG

$x \sqrt[]{3} = \frac{a}{2} - x + \frac{a}{2} - x$

x $\sqrt[]{3}$ = a – 2x

x = $\frac{a}{\sqrt[]{3} + 2}$

GE = x $\sqrt[]{3}$ = $\frac{a \sqrt[]{3}}{\sqrt[]{3} + 2}$

r = side of square ÷ 2

r= $\frac{a \sqrt[]{3}}{4 + 2 \sqrt[]{3}}$

ratio = a/r

= $\frac{a (4 + 2 \sqrt[]{3})}{a \sqrt[]{3}}$

= $\frac{4 + 2 \sqrt[]{3})}{\sqrt[]{3}} = \frac{4}{\sqrt[]{3}} + 2$

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7 months ago

Class 12th Physics

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential ( $V$ ) at any axial point, at 2 m distance ( $r$ ) from the centre of the dipole of dipole moment vector $\dot{P}$ of magnitude, $4 * 1 0^{- 6} C m$ , is $\pm 9 * 1 0^{3} V$ .

(Take $\frac{1}{4 π ϵ_{0}} = 9 * 1 0^{9}$ SI units)

Reason R : $V = \pm \frac{2 P}{4 π ?_{0} r^{2}}$ , where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below:

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alok kumar singh

Contributor-Level 10

The potential $V$ at any point, at distance $r$ from centre of dipole $= \frac{K P cos θ}{r^{2}}$

At axial point where $θ = 0^{?}, V = \frac{K P}{r^{2}} = \frac{9 * 1 0^{9} * 4 * 1 0^{- 6}}{2^{2}} = 9 * 1 0^{3} V$

At axial point where $θ = 18 0^{?}, V = \frac{- K P}{r^{2}} = - 9 * 1 0^{3} V$

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