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Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $\int_{0}^{1} [- 8 x^{2} + 6 x - 1] d x$ is equal to

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Payal Gupta

Contributor-Level 10

$\int_{0}^{1} [- 8 x^{2} + 6 x - 1] d x$

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) = $- \frac{D}{4 a} = - \frac{36 - 32}{- 32} = \frac{1}{8}$

= $- \frac{1}{4} - 1 (\frac{3}{4} - \frac{λ}{2}) - 2 (\frac{\sqrt[]{17} - 3}{8} - \frac{3}{4}) - 3 (1 - \frac{\sqrt[]{17} - 3}{8})$

= $\frac{\sqrt[]{17} - 13}{8}$

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

The magnetic field at the centre of current carrying circular loop is B₁. The magnetic field at a distance of $\sqrt[]{3}$ times radius of the given circular loop from the center on its axis is B₂. The value of B₁/B₂ will be

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Vishal Baghel

Contributor-Level 10

$B_{1} = \frac{μ_{0} i}{2 R}$

$B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + {(\sqrt[]{3} R)}^{2})}^{3 / 2}} = \frac{μ_{0} i R^{2}}{2 * 8 R^{3}} = \frac{μ_{0} i}{8 * 2 R} = \frac{B_{1}}{8}$

$\Rightarrow \frac{B_{1}}{B_{2}} = 8 : 1$

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Blocks of masses $m, 2 m$ and $8 m$ are arranged in a line on a frictionless floor. Another block of mass $m$ , moving with speed $v$ along the same line (see figure) collides with mass $m$ in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass $8 m$ starts moving the total energy loss is $p %$ of the original energy. Value of ' $p$ ' is close to:

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alok kumar singh

Contributor-Level 10

$? v$

$\Rightarrow m v = 16 {m v}_{1}$

$\Rightarrow V_{1} = \frac{V}{16}$

$\Rightarrow Δ k$ loss $= \frac{1}{2} {m v}^{2} - \frac{1}{2} (16 m) {(\frac{V}{16})}^{2}$

$\Rightarrow \frac{1}{2} {m v}^{2} (\frac{15}{16})$

$%$ loss $= \frac{15}{16} * 100 = 93.75 %$

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Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Let A₁, A₂, A₃,…. be an increasing geometric progression of positive real numbers. If A₁A₃A₅A₇ = $\frac{1}{1296}$ and A₂ + A₄ = $\frac{7}{36},$ then, the value of A₆ + A₈ + A₁₀ is equal to

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Payal Gupta

Contributor-Level 10

Let $\frac{A_{2}}{A_{1}} = \frac{A_{3}}{A_{2}} = . . . = r$

$∴ A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$

$\Rightarrow A_{1} r^{3} = \frac{1}{6} . . . . . . . (i)$

Again, A₂ + A₄ = $\frac{7}{36}$

$A_{1} r = \frac{7}{36} - \frac{1}{6} = \frac{1}{36} . . . . . . . . (i i)$

$(i) & (i i) r = \sqrt[]{6} & A_{1} = \frac{1}{36 \sqrt[]{6}}$

$∴ A_{6} + A_{8} + A_{10} = A_{1} r^{5} (1 + r^{2} + r^{4}) = 43$

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

A triangular shaped wire carrying 10 A current is a uniform magnetic field of 0.5T, as shown in figure.

The magnetic force on segment CD is (Given BC = CD = BD = 5cm).

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Vishal Baghel

Contributor-Level 10

$B = 0.5 T$

$F_{C D} = i \vec{l} * \vec{B}$

= $i l B s i n 60$

$= 10 * \frac{5}{100} * 0.5 * \frac{\sqrt[]{3}}{2} = \frac{\sqrt[]{3}}{8} = \frac{1.732}{8}$

= 0.216 N

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

Given below are two statements:

Statement I : A uniform wire of resistance 80 $Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 $Ω .$

Statement II : Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3 : 2.

In the light of the above statements, choose the most appropriate answer from the option given below

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Vishal Baghel

Contributor-Level 10

Statement – I

$80 = \frac{ρ l}{A}$

$S o, R_{e q} = \frac{R'}{4} = \frac{20}{4} = 5 Ω$

Statement – II

$\frac{P_{1}}{P_{2}} = \frac{V^{2}}{2 R} * \frac{3 R}{V^{2}} = \frac{3}{2}$

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Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is

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Payal Gupta

Contributor-Level 10

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = ²C₁ * 4! = 48

Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

Starting from the origin at time $t = 0$ , with initial velocity $5 \hat{j} {m s}^{- 1}$ , a particle moves in the $x$ $- y$ plane with a constant acceleration of $(10 \hat{i} + 4 \hat{j}) {m s}^{- 2}$ . At time $t$ , its coordinates are $(20$ $m, y_{0} m)$ . The values of $t$ and $y_{0}$ are, respectively:

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alok kumar singh

Contributor-Level 10

$\overset{?}{U} = 5 \hat{j}$

$\overset{?}{a} = 10 \hat{i} + 4 \hat{j}$

$\Rightarrow \overset{?}{S} = \overset{?}{U} t + \frac{1}{2} (\overset{?}{a}) t^{2}$

$\Rightarrow 20 \hat{i} + y_{0} \hat{j} = (5 t^{2}) \hat{i} + (5 t + 2 t^{2}) \hat{j}$

$\begin{array}{r} 20 = 5 t^{2} y_{0} = 5 t + 2 t^{2} \\ t = 2 \Rightarrow 18 m . \end{array}$

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Class 12th physics ncert exemplar solution class 12th chapter one

A uniform electric field E = (8m/e) V/m is created between two parallel plate of length 1m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s. The angle of the deviation ( θ) of the path of the electron as it comes out of the field will be________.

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Vishal Baghel

Contributor-Level 10

$E = \frac{8 m}{e} v / m$

Time to come out, t = $\frac{1}{2} s e c$

$V_{y} = a t = (\frac{e E}{m}) t = (\frac{e}{m} * \frac{8 m}{e}) * \frac{1}{2} = 4 m / s$

So, $t a n ? = \frac{V_{y}}{V_{x}} = \frac{4}{2}$

$? ? = t a n^{? 1} 2$

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New answer posted

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Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

Let A be matrix of order 3 * 3 and det(A) = 2. Then $d e t (d e t (A) a d j (5 a d j (A^{3})))$ is equal to

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Payal Gupta

Contributor-Level 10

|A| = 2

$| | A | a d j (5 a d j A^{3}) |$

= $| A P^{3} | | a d j (5 a d j (A^{3})) |$

$\begin{array}{l} = {| A |}^{15} . 5^{6} \\ = 2^{15} * 5^{6} = 2^{9} * 1 0^{6} \end{array}$

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