269593
The centre of mass of a non uniform rod of length \(L\) whose mass per unit length \(\lambda=\frac{K x^{2}}{L}\), Where \(\mathbf{k}\) is a constant and \(\mathbf{x}\) is the distance from one end is :
1 \(\frac{3 L}{4}\)
2 \(\frac{L}{8}\)
3 \(\frac{K}{L}\)
4 \(\frac{3 K}{L}\)
Explanation:
\(d m=\frac{k x^{2}}{L} d x ; x_{c m}=\frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m}=\frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x}=\frac{G^{4} L^{4}}{\square^{3} \square}=\frac{3 L}{4}\)
Rotational Motion
269594
A rope of length \(30 \mathrm{~cm}\) is on a horizontal table with maximum length hanging from edge \(A\) of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from \(A\) is
1 \(\frac{5 \sqrt{15}}{3} c\)
2 \(\frac{5 \sqrt{17}}{3} \mathrm{~cm}\)
3 \(\frac{5 \sqrt{19}}{3} \mathrm{~cm}\)
4 \(\frac{7 \sqrt{17}}{3} \mathrm{~cm}\)
Explanation:
Fractional length hanging,
\(\frac{l}{L}=\frac{\mu}{1+\mu} \Rightarrow \frac{l}{30}=\frac{0.5}{1+0.5} \Rightarrow l=10 \mathrm{~cm}\)
let ' \(\rho\) ' be the mass per unit length. The coordinates of \(20 \rho\) and \(10 \rho\) are \((10,0)\) and \((0,5)\) respectively from ' \(A\) '.
Distance of C.M from A, \(\mathrm{r}_{\mathrm{cm}}=\sqrt{X_{c m}^{2}+y_{c m}^{2}}\)
Rotational Motion
269595
As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is
1 \(4.2 \mathrm{~cm}\)
2 \(-4.2 \mathrm{~cm}\)
3 \(6.67 \mathrm{~cm}\)
4 \(-6.67 \mathrm{~cm}\)
Explanation:
\(\quad\) shift \(=\frac{- \text { mass of removed part } \times d}{\text { Mass of remaining part }}\) Here \(\mathrm{d}=20 \mathrm{~cm}\)
Rotational Motion
269596
A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumference of the discs coincide . The centre of mass of the new disc is \(\alpha R\) from the centre of the bigger disc. The value of \(\alpha\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 6\)
4 \(1 / 4\)
Explanation:
Shift of centre of mass \(\mathrm{x}=\frac{r^{2} a}{R^{2}-r^{2}}\)
Where \(r=\) radius of removed disc
\(\mathrm{R}=\) radius of original disc
\(\mathrm{a}=\) distance between the centres
Note:In this question shift must be \(\propto \mathrm{R}\) for exact approach to the solution
269593
The centre of mass of a non uniform rod of length \(L\) whose mass per unit length \(\lambda=\frac{K x^{2}}{L}\), Where \(\mathbf{k}\) is a constant and \(\mathbf{x}\) is the distance from one end is :
1 \(\frac{3 L}{4}\)
2 \(\frac{L}{8}\)
3 \(\frac{K}{L}\)
4 \(\frac{3 K}{L}\)
Explanation:
\(d m=\frac{k x^{2}}{L} d x ; x_{c m}=\frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m}=\frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x}=\frac{G^{4} L^{4}}{\square^{3} \square}=\frac{3 L}{4}\)
Rotational Motion
269594
A rope of length \(30 \mathrm{~cm}\) is on a horizontal table with maximum length hanging from edge \(A\) of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from \(A\) is
1 \(\frac{5 \sqrt{15}}{3} c\)
2 \(\frac{5 \sqrt{17}}{3} \mathrm{~cm}\)
3 \(\frac{5 \sqrt{19}}{3} \mathrm{~cm}\)
4 \(\frac{7 \sqrt{17}}{3} \mathrm{~cm}\)
Explanation:
Fractional length hanging,
\(\frac{l}{L}=\frac{\mu}{1+\mu} \Rightarrow \frac{l}{30}=\frac{0.5}{1+0.5} \Rightarrow l=10 \mathrm{~cm}\)
let ' \(\rho\) ' be the mass per unit length. The coordinates of \(20 \rho\) and \(10 \rho\) are \((10,0)\) and \((0,5)\) respectively from ' \(A\) '.
Distance of C.M from A, \(\mathrm{r}_{\mathrm{cm}}=\sqrt{X_{c m}^{2}+y_{c m}^{2}}\)
Rotational Motion
269595
As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is
1 \(4.2 \mathrm{~cm}\)
2 \(-4.2 \mathrm{~cm}\)
3 \(6.67 \mathrm{~cm}\)
4 \(-6.67 \mathrm{~cm}\)
Explanation:
\(\quad\) shift \(=\frac{- \text { mass of removed part } \times d}{\text { Mass of remaining part }}\) Here \(\mathrm{d}=20 \mathrm{~cm}\)
Rotational Motion
269596
A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumference of the discs coincide . The centre of mass of the new disc is \(\alpha R\) from the centre of the bigger disc. The value of \(\alpha\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 6\)
4 \(1 / 4\)
Explanation:
Shift of centre of mass \(\mathrm{x}=\frac{r^{2} a}{R^{2}-r^{2}}\)
Where \(r=\) radius of removed disc
\(\mathrm{R}=\) radius of original disc
\(\mathrm{a}=\) distance between the centres
Note:In this question shift must be \(\propto \mathrm{R}\) for exact approach to the solution
269593
The centre of mass of a non uniform rod of length \(L\) whose mass per unit length \(\lambda=\frac{K x^{2}}{L}\), Where \(\mathbf{k}\) is a constant and \(\mathbf{x}\) is the distance from one end is :
1 \(\frac{3 L}{4}\)
2 \(\frac{L}{8}\)
3 \(\frac{K}{L}\)
4 \(\frac{3 K}{L}\)
Explanation:
\(d m=\frac{k x^{2}}{L} d x ; x_{c m}=\frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m}=\frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x}=\frac{G^{4} L^{4}}{\square^{3} \square}=\frac{3 L}{4}\)
Rotational Motion
269594
A rope of length \(30 \mathrm{~cm}\) is on a horizontal table with maximum length hanging from edge \(A\) of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from \(A\) is
1 \(\frac{5 \sqrt{15}}{3} c\)
2 \(\frac{5 \sqrt{17}}{3} \mathrm{~cm}\)
3 \(\frac{5 \sqrt{19}}{3} \mathrm{~cm}\)
4 \(\frac{7 \sqrt{17}}{3} \mathrm{~cm}\)
Explanation:
Fractional length hanging,
\(\frac{l}{L}=\frac{\mu}{1+\mu} \Rightarrow \frac{l}{30}=\frac{0.5}{1+0.5} \Rightarrow l=10 \mathrm{~cm}\)
let ' \(\rho\) ' be the mass per unit length. The coordinates of \(20 \rho\) and \(10 \rho\) are \((10,0)\) and \((0,5)\) respectively from ' \(A\) '.
Distance of C.M from A, \(\mathrm{r}_{\mathrm{cm}}=\sqrt{X_{c m}^{2}+y_{c m}^{2}}\)
Rotational Motion
269595
As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is
1 \(4.2 \mathrm{~cm}\)
2 \(-4.2 \mathrm{~cm}\)
3 \(6.67 \mathrm{~cm}\)
4 \(-6.67 \mathrm{~cm}\)
Explanation:
\(\quad\) shift \(=\frac{- \text { mass of removed part } \times d}{\text { Mass of remaining part }}\) Here \(\mathrm{d}=20 \mathrm{~cm}\)
Rotational Motion
269596
A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumference of the discs coincide . The centre of mass of the new disc is \(\alpha R\) from the centre of the bigger disc. The value of \(\alpha\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 6\)
4 \(1 / 4\)
Explanation:
Shift of centre of mass \(\mathrm{x}=\frac{r^{2} a}{R^{2}-r^{2}}\)
Where \(r=\) radius of removed disc
\(\mathrm{R}=\) radius of original disc
\(\mathrm{a}=\) distance between the centres
Note:In this question shift must be \(\propto \mathrm{R}\) for exact approach to the solution
NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
269593
The centre of mass of a non uniform rod of length \(L\) whose mass per unit length \(\lambda=\frac{K x^{2}}{L}\), Where \(\mathbf{k}\) is a constant and \(\mathbf{x}\) is the distance from one end is :
1 \(\frac{3 L}{4}\)
2 \(\frac{L}{8}\)
3 \(\frac{K}{L}\)
4 \(\frac{3 K}{L}\)
Explanation:
\(d m=\frac{k x^{2}}{L} d x ; x_{c m}=\frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m}=\frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x}=\frac{G^{4} L^{4}}{\square^{3} \square}=\frac{3 L}{4}\)
Rotational Motion
269594
A rope of length \(30 \mathrm{~cm}\) is on a horizontal table with maximum length hanging from edge \(A\) of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from \(A\) is
1 \(\frac{5 \sqrt{15}}{3} c\)
2 \(\frac{5 \sqrt{17}}{3} \mathrm{~cm}\)
3 \(\frac{5 \sqrt{19}}{3} \mathrm{~cm}\)
4 \(\frac{7 \sqrt{17}}{3} \mathrm{~cm}\)
Explanation:
Fractional length hanging,
\(\frac{l}{L}=\frac{\mu}{1+\mu} \Rightarrow \frac{l}{30}=\frac{0.5}{1+0.5} \Rightarrow l=10 \mathrm{~cm}\)
let ' \(\rho\) ' be the mass per unit length. The coordinates of \(20 \rho\) and \(10 \rho\) are \((10,0)\) and \((0,5)\) respectively from ' \(A\) '.
Distance of C.M from A, \(\mathrm{r}_{\mathrm{cm}}=\sqrt{X_{c m}^{2}+y_{c m}^{2}}\)
Rotational Motion
269595
As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is
1 \(4.2 \mathrm{~cm}\)
2 \(-4.2 \mathrm{~cm}\)
3 \(6.67 \mathrm{~cm}\)
4 \(-6.67 \mathrm{~cm}\)
Explanation:
\(\quad\) shift \(=\frac{- \text { mass of removed part } \times d}{\text { Mass of remaining part }}\) Here \(\mathrm{d}=20 \mathrm{~cm}\)
Rotational Motion
269596
A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumference of the discs coincide . The centre of mass of the new disc is \(\alpha R\) from the centre of the bigger disc. The value of \(\alpha\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 6\)
4 \(1 / 4\)
Explanation:
Shift of centre of mass \(\mathrm{x}=\frac{r^{2} a}{R^{2}-r^{2}}\)
Where \(r=\) radius of removed disc
\(\mathrm{R}=\) radius of original disc
\(\mathrm{a}=\) distance between the centres
Note:In this question shift must be \(\propto \mathrm{R}\) for exact approach to the solution
