Let F- be the force acting on a particle having position vector r- and T- be the torque of this force about the origin- Then-

The correct option is D r⃗.T⃗=0 and F⃗.T⃗=0 T⃗=r⃗×F⃗ so, T⃗.r⃗=0 since T⃗⊥r⃗ so, T⃗.F⃗=0 since T⃗⊥F⃗ ...

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Byju's Answer

Standard XII

Physics

Acceleration in 2D

Let F⃗ be t...

Question

Let $\to F$ be the force acting on a particle having position vector $\to r$ and $\to T$ be the torque of this force about the origin. Then:

\to r . \to T = 0

and

\to F . \to T \neq 0

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\to r . \to T \neq 0

and

\to F . \to T = 0

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\to r . \to T \neq 0

and

\to F . \to T \neq 0

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\to r . \to T = 0

and

\to F . \to T = 0

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Solution

The correct option is D $\to r . \to T = 0$ and $\to F . \to T = 0$
$\to T = \to r \times \to F$
so, $\to T . \to r = 0$ since $\to T ⊥ \to r$
so, $\to T . \to F = 0$ since $\to T ⊥ \to F$

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Similar questions

Q. If

\to F

be a force acting on a particle having the position vector

\to r

and

\to τ

be the torque of this force about the origin, then,

\to τ . \to F = 0

and

\to r . \to τ = 0

Q. If

\to F

is the force acting on a particle having position vector

\to r

and

\to τ

be the torque of this force about the origin, then

Q. If

\to F

be a force acting on a particle having the position vector

\to r

and

\to τ

be the torque of this force about the origin, then

Q. Let

\underset{F}{\to}

be a force acting on a particle having position vector

\underset{r}{\to}

. Let

\underset{Γ}{\to}

be the torque of this force about the origin, then
(a)

\vec{r} . \vec{Γ} = 0 and \vec{F} . \vec{Γ} = 0

(b)

\vec{r} . \vec{Γ} = 0 but \vec{F} . \vec{Γ} \neq 0

(c)

\vec{r} . \vec{Γ} \neq 0 but \vec{F} . \vec{Γ} = 0

(d)

\vec{r} . \vec{Γ} \neq 0 and \vec{F} . \vec{Γ} \neq 0

Q. If

\to F

be a force acting on a particle having the position vector

\to r

and

\to τ

be the torque of this force about the origin, then

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Let →F be the force acting on a particle having position vector →r and →T be the torque of this force about the origin. Then:

The correct option is D →r.→T=0 and →F.→T=0→T=→r×→F so, →T.→r=0 since →T⊥→rso, →T.→F=0 since →T⊥→F

Let $\to F$ be the force acting on a particle having position vector $\to r$ and $\to T$ be the torque of this force about the origin. Then:

The correct option is D $\to r . \to T = 0$ and $\to F . \to T = 0$
$\to T = \to r \times \to F$
so, $\to T . \to r = 0$ since $\to T ⊥ \to r$
so, $\to T . \to F = 0$ since $\to T ⊥ \to F$