154232
The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 \mathrm{~cm}$ and $18 \mathrm{~cm}$ from the centre of a deflection magnetometer is
1 $2: 3$
2 $8: 27$
3 $27: 8$
4 $4: 9$
Explanation:
B Distance of first magnet $\left(\mathrm{d}_{1}\right)=12 \mathrm{~cm}$ Distance of second magnet $\left(\mathrm{d}_{2}\right)=18 \mathrm{~cm}$ $\Rightarrow \quad \mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^{3}}$ So, ratio of magnetic moment $\therefore \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\left(\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}\right)^{3}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{12}{18}\right)^{3}=\frac{8}{27}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\frac{8}{27}$ $\mathrm{M}_{1} : \mathrm{M}_{2}=8: 27$
Manipal UGET-2012
Magnetism and Matter
154233
The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{-4} \mathrm{~T}$ then magnetic filed at the centre of the coil due to current.
1 $1.96 \times 10^{-5} \mathrm{~T}$
2 $1.96 \times 10^{-4} \mathrm{~T}$
3 $1.96 \times 10^{4} \mathrm{~T}$
4 $1.96 \times 10^{5} \mathrm{~T}$
Explanation:
A Given that, Angle of deflection $(\theta)=30^{\circ}$ Horizontal magnetic field $\left(\mathrm{B}_{\mathrm{H}}\right)=0.34 \times 10^{-4} \mathrm{~T}$ Since we know that, Magnetic field $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta=0.34 \times 10^{-4} \times \tan 30^{\circ}$ $=0.34 \times 10^{-4} \times \frac{1}{\sqrt{3}}=0.196 \times 10^{-4}$ $=1.96 \times 10^{-5} \mathrm{~T}$
CG PET- 2008
Magnetism and Matter
154234
In a tangent galvanometer a current of $0.1 \mathrm{~A}$ produces a deflection of $30^{\circ}$. The current required to produce a deflection of $60^{\circ}$ will be
1 $0.6 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
D Given that, $i_{1}=0.1 \mathrm{~A}, \theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}$ We know that, magnetic field center $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta \quad\left(\because \mathrm{B}=\frac{\mu \mathrm{I}}{2 \mathrm{r}}\right)$ $\mathrm{I} \propto \tan \theta$ So, $\frac{I_{2}}{I_{1}}=\frac{\tan \theta_{2}}{\tan \theta_{1}}$ $\frac{\mathrm{I}_{2}}{0.1}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}$ $I_{2}=\frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$ $I_{2}=\frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$ $\mathrm{I}_{2}=0.3 \mathrm{~A}$
SRMJEEE - 2014
Magnetism and Matter
154236
A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $\mathbf{4 5}^{\circ}$. If the current be reduced by a factor of $\sqrt{3}$, the deflection would
1 decrease by $30^{\circ}$
2 decrease by $15^{\circ}$
3 increased by $15^{\circ}$
4 increased by $30^{\circ}$
Explanation:
B Given that, $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{\sqrt{3}}$ Angle of deflection $\left(\theta_{1}\right)=45^{\circ}$ We know that, $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta$ I $\propto \tan \theta$ Hence, $\frac{I_{1}}{I_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{1} / \sqrt{3}}=\frac{\tan 45^{\circ}}{\tan \theta_{2}}$ $\tan \theta_{2}=\frac{\tan 45^{\circ}}{\sqrt{3}}$ $\tan \theta_{2}=\frac{1}{\sqrt{3}}$ $\theta_{2}=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\theta_{2}=30^{\circ}$ Hence deflection decrease by, $\theta_{1}-\theta_{2}=45^{\circ}-30^{\circ}=15^{\circ}\left(\right.$ decrease by $\left.15^{\circ}\right)$
154232
The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 \mathrm{~cm}$ and $18 \mathrm{~cm}$ from the centre of a deflection magnetometer is
1 $2: 3$
2 $8: 27$
3 $27: 8$
4 $4: 9$
Explanation:
B Distance of first magnet $\left(\mathrm{d}_{1}\right)=12 \mathrm{~cm}$ Distance of second magnet $\left(\mathrm{d}_{2}\right)=18 \mathrm{~cm}$ $\Rightarrow \quad \mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^{3}}$ So, ratio of magnetic moment $\therefore \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\left(\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}\right)^{3}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{12}{18}\right)^{3}=\frac{8}{27}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\frac{8}{27}$ $\mathrm{M}_{1} : \mathrm{M}_{2}=8: 27$
Manipal UGET-2012
Magnetism and Matter
154233
The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{-4} \mathrm{~T}$ then magnetic filed at the centre of the coil due to current.
1 $1.96 \times 10^{-5} \mathrm{~T}$
2 $1.96 \times 10^{-4} \mathrm{~T}$
3 $1.96 \times 10^{4} \mathrm{~T}$
4 $1.96 \times 10^{5} \mathrm{~T}$
Explanation:
A Given that, Angle of deflection $(\theta)=30^{\circ}$ Horizontal magnetic field $\left(\mathrm{B}_{\mathrm{H}}\right)=0.34 \times 10^{-4} \mathrm{~T}$ Since we know that, Magnetic field $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta=0.34 \times 10^{-4} \times \tan 30^{\circ}$ $=0.34 \times 10^{-4} \times \frac{1}{\sqrt{3}}=0.196 \times 10^{-4}$ $=1.96 \times 10^{-5} \mathrm{~T}$
CG PET- 2008
Magnetism and Matter
154234
In a tangent galvanometer a current of $0.1 \mathrm{~A}$ produces a deflection of $30^{\circ}$. The current required to produce a deflection of $60^{\circ}$ will be
1 $0.6 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
D Given that, $i_{1}=0.1 \mathrm{~A}, \theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}$ We know that, magnetic field center $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta \quad\left(\because \mathrm{B}=\frac{\mu \mathrm{I}}{2 \mathrm{r}}\right)$ $\mathrm{I} \propto \tan \theta$ So, $\frac{I_{2}}{I_{1}}=\frac{\tan \theta_{2}}{\tan \theta_{1}}$ $\frac{\mathrm{I}_{2}}{0.1}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}$ $I_{2}=\frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$ $I_{2}=\frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$ $\mathrm{I}_{2}=0.3 \mathrm{~A}$
SRMJEEE - 2014
Magnetism and Matter
154236
A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $\mathbf{4 5}^{\circ}$. If the current be reduced by a factor of $\sqrt{3}$, the deflection would
1 decrease by $30^{\circ}$
2 decrease by $15^{\circ}$
3 increased by $15^{\circ}$
4 increased by $30^{\circ}$
Explanation:
B Given that, $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{\sqrt{3}}$ Angle of deflection $\left(\theta_{1}\right)=45^{\circ}$ We know that, $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta$ I $\propto \tan \theta$ Hence, $\frac{I_{1}}{I_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{1} / \sqrt{3}}=\frac{\tan 45^{\circ}}{\tan \theta_{2}}$ $\tan \theta_{2}=\frac{\tan 45^{\circ}}{\sqrt{3}}$ $\tan \theta_{2}=\frac{1}{\sqrt{3}}$ $\theta_{2}=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\theta_{2}=30^{\circ}$ Hence deflection decrease by, $\theta_{1}-\theta_{2}=45^{\circ}-30^{\circ}=15^{\circ}\left(\right.$ decrease by $\left.15^{\circ}\right)$
154232
The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 \mathrm{~cm}$ and $18 \mathrm{~cm}$ from the centre of a deflection magnetometer is
1 $2: 3$
2 $8: 27$
3 $27: 8$
4 $4: 9$
Explanation:
B Distance of first magnet $\left(\mathrm{d}_{1}\right)=12 \mathrm{~cm}$ Distance of second magnet $\left(\mathrm{d}_{2}\right)=18 \mathrm{~cm}$ $\Rightarrow \quad \mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^{3}}$ So, ratio of magnetic moment $\therefore \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\left(\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}\right)^{3}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{12}{18}\right)^{3}=\frac{8}{27}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\frac{8}{27}$ $\mathrm{M}_{1} : \mathrm{M}_{2}=8: 27$
Manipal UGET-2012
Magnetism and Matter
154233
The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{-4} \mathrm{~T}$ then magnetic filed at the centre of the coil due to current.
1 $1.96 \times 10^{-5} \mathrm{~T}$
2 $1.96 \times 10^{-4} \mathrm{~T}$
3 $1.96 \times 10^{4} \mathrm{~T}$
4 $1.96 \times 10^{5} \mathrm{~T}$
Explanation:
A Given that, Angle of deflection $(\theta)=30^{\circ}$ Horizontal magnetic field $\left(\mathrm{B}_{\mathrm{H}}\right)=0.34 \times 10^{-4} \mathrm{~T}$ Since we know that, Magnetic field $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta=0.34 \times 10^{-4} \times \tan 30^{\circ}$ $=0.34 \times 10^{-4} \times \frac{1}{\sqrt{3}}=0.196 \times 10^{-4}$ $=1.96 \times 10^{-5} \mathrm{~T}$
CG PET- 2008
Magnetism and Matter
154234
In a tangent galvanometer a current of $0.1 \mathrm{~A}$ produces a deflection of $30^{\circ}$. The current required to produce a deflection of $60^{\circ}$ will be
1 $0.6 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
D Given that, $i_{1}=0.1 \mathrm{~A}, \theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}$ We know that, magnetic field center $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta \quad\left(\because \mathrm{B}=\frac{\mu \mathrm{I}}{2 \mathrm{r}}\right)$ $\mathrm{I} \propto \tan \theta$ So, $\frac{I_{2}}{I_{1}}=\frac{\tan \theta_{2}}{\tan \theta_{1}}$ $\frac{\mathrm{I}_{2}}{0.1}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}$ $I_{2}=\frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$ $I_{2}=\frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$ $\mathrm{I}_{2}=0.3 \mathrm{~A}$
SRMJEEE - 2014
Magnetism and Matter
154236
A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $\mathbf{4 5}^{\circ}$. If the current be reduced by a factor of $\sqrt{3}$, the deflection would
1 decrease by $30^{\circ}$
2 decrease by $15^{\circ}$
3 increased by $15^{\circ}$
4 increased by $30^{\circ}$
Explanation:
B Given that, $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{\sqrt{3}}$ Angle of deflection $\left(\theta_{1}\right)=45^{\circ}$ We know that, $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta$ I $\propto \tan \theta$ Hence, $\frac{I_{1}}{I_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{1} / \sqrt{3}}=\frac{\tan 45^{\circ}}{\tan \theta_{2}}$ $\tan \theta_{2}=\frac{\tan 45^{\circ}}{\sqrt{3}}$ $\tan \theta_{2}=\frac{1}{\sqrt{3}}$ $\theta_{2}=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\theta_{2}=30^{\circ}$ Hence deflection decrease by, $\theta_{1}-\theta_{2}=45^{\circ}-30^{\circ}=15^{\circ}\left(\right.$ decrease by $\left.15^{\circ}\right)$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Magnetism and Matter
154232
The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 \mathrm{~cm}$ and $18 \mathrm{~cm}$ from the centre of a deflection magnetometer is
1 $2: 3$
2 $8: 27$
3 $27: 8$
4 $4: 9$
Explanation:
B Distance of first magnet $\left(\mathrm{d}_{1}\right)=12 \mathrm{~cm}$ Distance of second magnet $\left(\mathrm{d}_{2}\right)=18 \mathrm{~cm}$ $\Rightarrow \quad \mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^{3}}$ So, ratio of magnetic moment $\therefore \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\left(\frac{\mathrm{d}_{1}}{\mathrm{~d}_{2}}\right)^{3}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{12}{18}\right)^{3}=\frac{8}{27}$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}} =\frac{8}{27}$ $\mathrm{M}_{1} : \mathrm{M}_{2}=8: 27$
Manipal UGET-2012
Magnetism and Matter
154233
The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{-4} \mathrm{~T}$ then magnetic filed at the centre of the coil due to current.
1 $1.96 \times 10^{-5} \mathrm{~T}$
2 $1.96 \times 10^{-4} \mathrm{~T}$
3 $1.96 \times 10^{4} \mathrm{~T}$
4 $1.96 \times 10^{5} \mathrm{~T}$
Explanation:
A Given that, Angle of deflection $(\theta)=30^{\circ}$ Horizontal magnetic field $\left(\mathrm{B}_{\mathrm{H}}\right)=0.34 \times 10^{-4} \mathrm{~T}$ Since we know that, Magnetic field $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta=0.34 \times 10^{-4} \times \tan 30^{\circ}$ $=0.34 \times 10^{-4} \times \frac{1}{\sqrt{3}}=0.196 \times 10^{-4}$ $=1.96 \times 10^{-5} \mathrm{~T}$
CG PET- 2008
Magnetism and Matter
154234
In a tangent galvanometer a current of $0.1 \mathrm{~A}$ produces a deflection of $30^{\circ}$. The current required to produce a deflection of $60^{\circ}$ will be
1 $0.6 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $0.3 \mathrm{~A}$
Explanation:
D Given that, $i_{1}=0.1 \mathrm{~A}, \theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}$ We know that, magnetic field center $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta \quad\left(\because \mathrm{B}=\frac{\mu \mathrm{I}}{2 \mathrm{r}}\right)$ $\mathrm{I} \propto \tan \theta$ So, $\frac{I_{2}}{I_{1}}=\frac{\tan \theta_{2}}{\tan \theta_{1}}$ $\frac{\mathrm{I}_{2}}{0.1}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}$ $I_{2}=\frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$ $I_{2}=\frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$ $\mathrm{I}_{2}=0.3 \mathrm{~A}$
SRMJEEE - 2014
Magnetism and Matter
154236
A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $\mathbf{4 5}^{\circ}$. If the current be reduced by a factor of $\sqrt{3}$, the deflection would
1 decrease by $30^{\circ}$
2 decrease by $15^{\circ}$
3 increased by $15^{\circ}$
4 increased by $30^{\circ}$
Explanation:
B Given that, $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{\sqrt{3}}$ Angle of deflection $\left(\theta_{1}\right)=45^{\circ}$ We know that, $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\mathrm{B}_{\mathrm{H}} \tan \theta$ I $\propto \tan \theta$ Hence, $\frac{I_{1}}{I_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{1} / \sqrt{3}}=\frac{\tan 45^{\circ}}{\tan \theta_{2}}$ $\tan \theta_{2}=\frac{\tan 45^{\circ}}{\sqrt{3}}$ $\tan \theta_{2}=\frac{1}{\sqrt{3}}$ $\theta_{2}=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\theta_{2}=30^{\circ}$ Hence deflection decrease by, $\theta_{1}-\theta_{2}=45^{\circ}-30^{\circ}=15^{\circ}\left(\right.$ decrease by $\left.15^{\circ}\right)$
