Class 11th

Class 11th System of Particles and Rotational Motion

Contributor-Level 10

$g^{1} = \frac{G M}{{(R + h)}^{2}}$

$g^{1} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

Given h = D = 2R

$g^{1} = \frac{g}{{(1 + \frac{2 R}{R})}^{2}}$

$g^{1} = \frac{g}{9}$

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New answer posted

3 months ago

If force $\vec{F} = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$ acts on a particle having position vector $2 \hat{i} + \hat{j} + 2 \hat{k}$ then, the torque about the origin will be:

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Class 11th Motion in a Plane

Contributor-Level 10

^{$\vec{r} = 2 \hat{i} + \hat{j} + 2 \hat{k}$}

^{$\vec{τ} = \vec{r} * \vec{F}$}

^{$= (2 \hat{i} + \hat{j} + 2 \hat{k}) * (3 \hat{i} + 4 \hat{j} - 2 \hat{k}) = | \begin{matrix} \hat{i} & - \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & - 2 \end{matrix} |$}

^{$\vec{τ} = - 10 \hat{i} + 10 \hat{j} + 5 \hat{k}$}

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New answer posted

3 months ago

Which one of the following relations is true for two unit vector $\hat{A} a n d \hat{B}$ making an angle θ to each other?

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Class 11th Motion in a Plane

Contributor-Level 10

$| \hat{A} + \hat{B} | = \sqrt[]{A^{2} + B^{2} + 2 A B c o s θ}$ $^{}^{}$

$= \sqrt[]{1 + 1 + 2 c o s θ}$

$= \sqrt[]{2 (1 + c o s θ)}$

= $\sqrt[]{2 * 2 c o s^{2} \frac{θ}{2}}$ $^{} \frac{}{}$

$| \hat{A} + \hat{B} | = 2 c o s \frac{θ}{2}$ -(1)

$| \hat{A} - \hat{B} | = \sqrt[]{A^{2} + B^{2} - 2 A B c o s θ}$

$= \sqrt[]{1 + 1 - 2 c o s θ}$

$= 2 s i n \frac{θ}{2}$ -(2)

(2) $\div$ (1)

$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 s i n \frac{θ}{2}}{2 c o s \frac{θ}{2}}$

$\Rightarrow | \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | t a n \frac{θ}{2}$

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New answer posted

3 months ago

$\vec{A}$ is a vector quantity such that $| \vec{A} |$ = non-zero constant. Which of the following expression is true for $\vec{A}$ ?

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Class 11th Units and Measurement

Contributor-Level 10

$\vec{A} * \vec{A} = A A s i n θ \hat{n}$

$= A A s i n 0 ° \hat{n}$

= 0 [since Angle between the vectors are zero degree]

$\vec{A} * \vec{A} = 0$

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New answer posted

3 months ago

If $Z = \frac{A^{2} B^{3}}{C^{4}},$ then the relative error in Z will be:

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Contributor-Level 10

$z = \frac{A^{2} B^{3}}{C^{4}}$

The relative error in z is given by

$\frac{Δ z}{z} = \frac{2 Δ A}{A} + \frac{3 Δ B}{B} + \frac{4 Δ C}{C}$

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3 months ago

The statement $(p \Rightarrow q) \lor (p \Rightarrow r)$ is NOT equivalent to

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Class 11th Relations and Functions

Contributor-Level 10

$(A) \sim (p \land (\sim r)) \lor q = (\sim p) \lor (r) \lor q$

(B) $q \lor (- r \lor p)$

(C) $(\sim p) \lor (q \lor r)$

(D) $(\sim p \lor q) \lor r$

Using Venn diagram we get B as the correct option.

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3 months ago

The domain of the function f(x) = $s i n^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) i s :$

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Contributor-Level 10

$- 1 \leq \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$

$\Rightarrow 0 \leq \frac{2 x^{2} - x + 9}{x^{2} + 2 x + 7} & \frac{- 5 x - 5}{x^{2} + 2 x + 7} \leq 0$

$x \in R & - 1 \leq x < \infty$

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3 months ago

If (2, 3, 9), (5, 2, 1), $(1, λ, 8)$ and $(λ, 2, 3)$ are coplanar, then the product of all possible values of $λ$ is

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Contributor-Level 10

A(2, 3, 9)

B(5, 2, 1)

C(1, $λ$ , 8)

D( $λ$ ,2, 3)

$Δ = | \begin{matrix} 3 & - 1 & - 8 \\ - 4 & λ - 2 & 7 \\ λ - 2 & - 1 & - 6 \end{matrix} |$

$\vec{A B} = (3, - 1, - 8)$

$\vec{A C} = (- 4, λ - 2, 7)$

$\vec{A D} = (λ - 2, - 1, - 6)$

$Δ = 3 [- 6 λ + 12 + 7] - 1 (7 λ - 14 - 24) - 8 (4 - {(λ - 2)}^{2})$

$= 57 - 18 λ \Rightarrow + 38 - 32 + 8 (λ^{2} - 4 λ + 4)$

$= 95 - 57 λ + 8 λ^{2}$

$Δ = 0 \Rightarrow λ_{1} λ_{2} = \frac{95}{8}$

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New answer posted

3 months ago

Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that $∠ P R Q = 60 °,$ then the area of $Δ P Q R$ is equal to

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Contributor-Level 10

x + 2y + z = 14

$⊥^{r} l i n e P Q : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{1} = t$

Q (1 + t, 2 + 2t, 3 + t)

x + 2y + z = 14 ⇒ 1 + t + 4 + 4t + 3 + t = 14 ⇒ 6t = 6

t = 1

⇒ Q (2, 4, 4)

PQ = $\sqrt[]{1 + 4 + 1} = \sqrt[]{6}$

$t a n 60 ° = \frac{P Q}{Q R} \Rightarrow Q R = \frac{P Q}{\sqrt[]{3}} = \sqrt[]{2}$

ar (PQR) = $\frac{1}{2} \sqrt[]{6} \cdot \sqrt[]{2} = \sqrt[]{3}$

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3 months ago

Let A(, 2), B(, 6) and $C (\frac{α}{4}, - 2)$ be vertices of a $Δ A B C .$ If $(5, \frac{α}{4})$ is the circumcentre of $Δ A B C,$ then which of the following is NOT correct about $Δ A B C ?$

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