355559
The resultant of two forces at right angle is 5 \(N\). When the angle between them is \(120^{\circ}\), the resultant is \(\sqrt{13}\). Then, the force is
1 \(\sqrt{20} N, \sqrt{5} N\)
2 \(\sqrt{12} N, \sqrt{13} N\)
3 \(\sqrt{40} N, \sqrt{15} N\)
4 \(3 N, 4 N\)
Explanation:
Let, \(A\) and \(B\) be the two forces. \(\sqrt{A^{2}+B^{2}}=5\) or \({A^2} + {B^2} = 25\,\,\,\,\,\,(1)\) and \(A^{2}+B^{2}+2 A B \cos 120^{\circ}=13\) or \(25+2 A B \times(-1 / 2)=13\) or \(2{\rm{ }}AB = 24\,\,\,\,\,\,\,\,\,\,\,(2)\) Solving eq's (1) and (2), we get \(A=3 N\) and \(B=4 N\)
PHXI06:WORK ENERGY AND POWER
355560
Given \(\vec{R}=\vec{A}+\vec{B}\) and \(R=A=B\). the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(90^{\circ}\)
2 \(60^{\circ}\)
3 \(180^{\circ}\)
4 \(120^{\circ}\)
Explanation:
For the resultant, \(\begin{aligned}& R^{2}=R^{2}+R^{2}+2 R^{2} \cos \theta \\& \Rightarrow R^{2}=2 R^{2}+2 R^{2} \cos \theta \\& \dfrac{1}{2}=1+\cos \theta\end{aligned}\) or \(\cos \theta=-\dfrac{1}{2}\) or \(\theta=120^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355561
Given that \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{C}\) is \(\perp\) to \(\vec{A}\). Further if \(|\vec{A}|=|\vec{C}|\), then what is the angle between \(\vec{A}\) and \(\vec{B}\)
1 \(\dfrac{\pi}{4}\) radian
2 \(\dfrac{\pi}{2}\)
3 \(\dfrac{3 \pi}{4}\) radian
4 \(\pi\) radian
Explanation:
Here, \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{B}=\vec{C}-\vec{A}\) Since \(\vec{C} \perp \vec{A}\) therefore \(B^{2}=C^{2}+A^{2}-2 C A \cos \left(90^{\circ}\right)\) Also, \(|\vec{C}|=|\vec{A}|\) \(B^{2}=2 A^{2} \Rightarrow B=\sqrt{2} A\) Now, \(A^{2}+B^{2}+2 A B \cos \theta=C^{2}=A^{2}\) Therefore we find \(\cos \theta=-\dfrac{1}{\sqrt{2}}\) This gives \(\theta=\dfrac{3 \pi}{4}\)
PHXI06:WORK ENERGY AND POWER
355562
If \(A B \cos \theta=A B\), then the angle between \(\mathrm{A}\) and \(\mathrm{B}\) is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
As, \(A B \cos \theta=A B\) \(\cos \theta=1 \Rightarrow \theta=0^{\circ}\)
355559
The resultant of two forces at right angle is 5 \(N\). When the angle between them is \(120^{\circ}\), the resultant is \(\sqrt{13}\). Then, the force is
1 \(\sqrt{20} N, \sqrt{5} N\)
2 \(\sqrt{12} N, \sqrt{13} N\)
3 \(\sqrt{40} N, \sqrt{15} N\)
4 \(3 N, 4 N\)
Explanation:
Let, \(A\) and \(B\) be the two forces. \(\sqrt{A^{2}+B^{2}}=5\) or \({A^2} + {B^2} = 25\,\,\,\,\,\,(1)\) and \(A^{2}+B^{2}+2 A B \cos 120^{\circ}=13\) or \(25+2 A B \times(-1 / 2)=13\) or \(2{\rm{ }}AB = 24\,\,\,\,\,\,\,\,\,\,\,(2)\) Solving eq's (1) and (2), we get \(A=3 N\) and \(B=4 N\)
PHXI06:WORK ENERGY AND POWER
355560
Given \(\vec{R}=\vec{A}+\vec{B}\) and \(R=A=B\). the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(90^{\circ}\)
2 \(60^{\circ}\)
3 \(180^{\circ}\)
4 \(120^{\circ}\)
Explanation:
For the resultant, \(\begin{aligned}& R^{2}=R^{2}+R^{2}+2 R^{2} \cos \theta \\& \Rightarrow R^{2}=2 R^{2}+2 R^{2} \cos \theta \\& \dfrac{1}{2}=1+\cos \theta\end{aligned}\) or \(\cos \theta=-\dfrac{1}{2}\) or \(\theta=120^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355561
Given that \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{C}\) is \(\perp\) to \(\vec{A}\). Further if \(|\vec{A}|=|\vec{C}|\), then what is the angle between \(\vec{A}\) and \(\vec{B}\)
1 \(\dfrac{\pi}{4}\) radian
2 \(\dfrac{\pi}{2}\)
3 \(\dfrac{3 \pi}{4}\) radian
4 \(\pi\) radian
Explanation:
Here, \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{B}=\vec{C}-\vec{A}\) Since \(\vec{C} \perp \vec{A}\) therefore \(B^{2}=C^{2}+A^{2}-2 C A \cos \left(90^{\circ}\right)\) Also, \(|\vec{C}|=|\vec{A}|\) \(B^{2}=2 A^{2} \Rightarrow B=\sqrt{2} A\) Now, \(A^{2}+B^{2}+2 A B \cos \theta=C^{2}=A^{2}\) Therefore we find \(\cos \theta=-\dfrac{1}{\sqrt{2}}\) This gives \(\theta=\dfrac{3 \pi}{4}\)
PHXI06:WORK ENERGY AND POWER
355562
If \(A B \cos \theta=A B\), then the angle between \(\mathrm{A}\) and \(\mathrm{B}\) is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
As, \(A B \cos \theta=A B\) \(\cos \theta=1 \Rightarrow \theta=0^{\circ}\)
355559
The resultant of two forces at right angle is 5 \(N\). When the angle between them is \(120^{\circ}\), the resultant is \(\sqrt{13}\). Then, the force is
1 \(\sqrt{20} N, \sqrt{5} N\)
2 \(\sqrt{12} N, \sqrt{13} N\)
3 \(\sqrt{40} N, \sqrt{15} N\)
4 \(3 N, 4 N\)
Explanation:
Let, \(A\) and \(B\) be the two forces. \(\sqrt{A^{2}+B^{2}}=5\) or \({A^2} + {B^2} = 25\,\,\,\,\,\,(1)\) and \(A^{2}+B^{2}+2 A B \cos 120^{\circ}=13\) or \(25+2 A B \times(-1 / 2)=13\) or \(2{\rm{ }}AB = 24\,\,\,\,\,\,\,\,\,\,\,(2)\) Solving eq's (1) and (2), we get \(A=3 N\) and \(B=4 N\)
PHXI06:WORK ENERGY AND POWER
355560
Given \(\vec{R}=\vec{A}+\vec{B}\) and \(R=A=B\). the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(90^{\circ}\)
2 \(60^{\circ}\)
3 \(180^{\circ}\)
4 \(120^{\circ}\)
Explanation:
For the resultant, \(\begin{aligned}& R^{2}=R^{2}+R^{2}+2 R^{2} \cos \theta \\& \Rightarrow R^{2}=2 R^{2}+2 R^{2} \cos \theta \\& \dfrac{1}{2}=1+\cos \theta\end{aligned}\) or \(\cos \theta=-\dfrac{1}{2}\) or \(\theta=120^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355561
Given that \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{C}\) is \(\perp\) to \(\vec{A}\). Further if \(|\vec{A}|=|\vec{C}|\), then what is the angle between \(\vec{A}\) and \(\vec{B}\)
1 \(\dfrac{\pi}{4}\) radian
2 \(\dfrac{\pi}{2}\)
3 \(\dfrac{3 \pi}{4}\) radian
4 \(\pi\) radian
Explanation:
Here, \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{B}=\vec{C}-\vec{A}\) Since \(\vec{C} \perp \vec{A}\) therefore \(B^{2}=C^{2}+A^{2}-2 C A \cos \left(90^{\circ}\right)\) Also, \(|\vec{C}|=|\vec{A}|\) \(B^{2}=2 A^{2} \Rightarrow B=\sqrt{2} A\) Now, \(A^{2}+B^{2}+2 A B \cos \theta=C^{2}=A^{2}\) Therefore we find \(\cos \theta=-\dfrac{1}{\sqrt{2}}\) This gives \(\theta=\dfrac{3 \pi}{4}\)
PHXI06:WORK ENERGY AND POWER
355562
If \(A B \cos \theta=A B\), then the angle between \(\mathrm{A}\) and \(\mathrm{B}\) is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
As, \(A B \cos \theta=A B\) \(\cos \theta=1 \Rightarrow \theta=0^{\circ}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI06:WORK ENERGY AND POWER
355559
The resultant of two forces at right angle is 5 \(N\). When the angle between them is \(120^{\circ}\), the resultant is \(\sqrt{13}\). Then, the force is
1 \(\sqrt{20} N, \sqrt{5} N\)
2 \(\sqrt{12} N, \sqrt{13} N\)
3 \(\sqrt{40} N, \sqrt{15} N\)
4 \(3 N, 4 N\)
Explanation:
Let, \(A\) and \(B\) be the two forces. \(\sqrt{A^{2}+B^{2}}=5\) or \({A^2} + {B^2} = 25\,\,\,\,\,\,(1)\) and \(A^{2}+B^{2}+2 A B \cos 120^{\circ}=13\) or \(25+2 A B \times(-1 / 2)=13\) or \(2{\rm{ }}AB = 24\,\,\,\,\,\,\,\,\,\,\,(2)\) Solving eq's (1) and (2), we get \(A=3 N\) and \(B=4 N\)
PHXI06:WORK ENERGY AND POWER
355560
Given \(\vec{R}=\vec{A}+\vec{B}\) and \(R=A=B\). the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(90^{\circ}\)
2 \(60^{\circ}\)
3 \(180^{\circ}\)
4 \(120^{\circ}\)
Explanation:
For the resultant, \(\begin{aligned}& R^{2}=R^{2}+R^{2}+2 R^{2} \cos \theta \\& \Rightarrow R^{2}=2 R^{2}+2 R^{2} \cos \theta \\& \dfrac{1}{2}=1+\cos \theta\end{aligned}\) or \(\cos \theta=-\dfrac{1}{2}\) or \(\theta=120^{\circ}\)
PHXI06:WORK ENERGY AND POWER
355561
Given that \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{C}\) is \(\perp\) to \(\vec{A}\). Further if \(|\vec{A}|=|\vec{C}|\), then what is the angle between \(\vec{A}\) and \(\vec{B}\)
1 \(\dfrac{\pi}{4}\) radian
2 \(\dfrac{\pi}{2}\)
3 \(\dfrac{3 \pi}{4}\) radian
4 \(\pi\) radian
Explanation:
Here, \(\vec{A}+\vec{B}=\vec{C}\) and \(\vec{B}=\vec{C}-\vec{A}\) Since \(\vec{C} \perp \vec{A}\) therefore \(B^{2}=C^{2}+A^{2}-2 C A \cos \left(90^{\circ}\right)\) Also, \(|\vec{C}|=|\vec{A}|\) \(B^{2}=2 A^{2} \Rightarrow B=\sqrt{2} A\) Now, \(A^{2}+B^{2}+2 A B \cos \theta=C^{2}=A^{2}\) Therefore we find \(\cos \theta=-\dfrac{1}{\sqrt{2}}\) This gives \(\theta=\dfrac{3 \pi}{4}\)
PHXI06:WORK ENERGY AND POWER
355562
If \(A B \cos \theta=A B\), then the angle between \(\mathrm{A}\) and \(\mathrm{B}\) is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(90^{\circ}\)
4 \(180^{\circ}\)
Explanation:
As, \(A B \cos \theta=A B\) \(\cos \theta=1 \Rightarrow \theta=0^{\circ}\)