NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355538
Assertion : The scalar product of two vectors can be zero. Reason : If two vectors are perpendicular to each other, their scalar product will be zero.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(W=F S \cos \theta\), the scalar product of two vectors can indeed be zero. This happens when two vectors are perpendicular to each other as \(\cos 90^{\circ}=0\), in which case their scalar product will be zero. So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355539
If \(\vec{A}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{B}=-\hat{i}+3 \hat{j}+4 \hat{k}\) then projection of \(\vec{A}\) on \(\vec{B}\) will be
1 \(\dfrac{3}{\sqrt{13}}\)
2 \(\dfrac{3}{\sqrt{26}}\)
3 \(\sqrt{\dfrac{3}{26}}\)
4 \(\sqrt{\dfrac{3}{13}}\)
Explanation:
The projection of \(\vec{A}\) on \(\vec{B}=\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}\) \(\begin{gathered}|\vec{B}|=\sqrt{(-1)^{2}+3^{2}+4^{2}}=\sqrt{1+9+16}=\sqrt{26} \\\vec{A} \cdot \vec{B}=2(-1)+3 \times 3+(-1)(4)=3 \\\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}=\dfrac{3}{\sqrt{26}}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355540
If the sum of two unit vectors is a unit vector, then magnitude of difference is
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(1 / \sqrt{2}\)
4 \(\sqrt{5}\)
Explanation:
Let \(\hat{n}_{1}\) and \(\hat{n}_{2}\) are the two unit vectors, then the sum is \(\vec{n}_{s}=\hat{n}_{1}+\hat{n}_{2}\) or \(n_{s}^{2}=n_{1}^{2}+n_{2}^{2}+2 n_{1} n_{2} \cos \theta\) \(=1+1+2 \cos \theta\) Since it is given that \(n_{s}\) is also a unit vector, therefore \(1=1+1+2 \cos \theta \Rightarrow \cos \theta=-\dfrac{1}{2} \therefore \theta=120^{\circ}\) Now the difference vector is \(\hat{n}_{d}=\hat{n}_{1}-\hat{n}_{2}\) or \(\begin{aligned}& n_{d}^{2}=n_{1}^{2}+n_{2}^{2}-2 n_{1} n_{2} \cos \theta=1+1-2 \cos \left(120^{\circ}\right) \\& \quad \therefore n_{d}^{2}=2-2(-1 / 2)=2+1=3 \Rightarrow n_{d}=\sqrt{3}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355541
If \(\vec{a}+\vec{b}=\vec{c}\) and \(a+b=c\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
1 \(90^{\circ}\)
2 \(180^{\circ}\)
3 \(120^{\circ}\)
4 Zero
Explanation:
Here, \(\vec{a}+\vec{b}=\vec{c}\) and \(c=a+b\) Let \(\theta\) be angle between \(\vec{a}\) and \(\vec{b}\). \(\begin{aligned}& \therefore c=\sqrt{a^{2}+b^{2}+2 a b \cos \theta} \\& a+b=\sqrt{a^{2}+b^{2}+2 a b \cos \theta}\end{aligned}\) Squaring both sides, we get \(\begin{aligned}& (a+b)^{2}=a^{2}+b^{2}+2 a b \cos \theta \\& a^{2}+b^{2}+2 a b=a^{2}+b^{2}+2 a b \cos \theta \\& 2 a b-2 a b \cos \theta=0 \text { or } 1-\cos \theta=0 \\& \cos \theta=1 \text { or } \theta=\cos ^{-1}(1)=0^{0} .\end{aligned}\)
355538
Assertion : The scalar product of two vectors can be zero. Reason : If two vectors are perpendicular to each other, their scalar product will be zero.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(W=F S \cos \theta\), the scalar product of two vectors can indeed be zero. This happens when two vectors are perpendicular to each other as \(\cos 90^{\circ}=0\), in which case their scalar product will be zero. So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355539
If \(\vec{A}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{B}=-\hat{i}+3 \hat{j}+4 \hat{k}\) then projection of \(\vec{A}\) on \(\vec{B}\) will be
1 \(\dfrac{3}{\sqrt{13}}\)
2 \(\dfrac{3}{\sqrt{26}}\)
3 \(\sqrt{\dfrac{3}{26}}\)
4 \(\sqrt{\dfrac{3}{13}}\)
Explanation:
The projection of \(\vec{A}\) on \(\vec{B}=\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}\) \(\begin{gathered}|\vec{B}|=\sqrt{(-1)^{2}+3^{2}+4^{2}}=\sqrt{1+9+16}=\sqrt{26} \\\vec{A} \cdot \vec{B}=2(-1)+3 \times 3+(-1)(4)=3 \\\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}=\dfrac{3}{\sqrt{26}}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355540
If the sum of two unit vectors is a unit vector, then magnitude of difference is
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(1 / \sqrt{2}\)
4 \(\sqrt{5}\)
Explanation:
Let \(\hat{n}_{1}\) and \(\hat{n}_{2}\) are the two unit vectors, then the sum is \(\vec{n}_{s}=\hat{n}_{1}+\hat{n}_{2}\) or \(n_{s}^{2}=n_{1}^{2}+n_{2}^{2}+2 n_{1} n_{2} \cos \theta\) \(=1+1+2 \cos \theta\) Since it is given that \(n_{s}\) is also a unit vector, therefore \(1=1+1+2 \cos \theta \Rightarrow \cos \theta=-\dfrac{1}{2} \therefore \theta=120^{\circ}\) Now the difference vector is \(\hat{n}_{d}=\hat{n}_{1}-\hat{n}_{2}\) or \(\begin{aligned}& n_{d}^{2}=n_{1}^{2}+n_{2}^{2}-2 n_{1} n_{2} \cos \theta=1+1-2 \cos \left(120^{\circ}\right) \\& \quad \therefore n_{d}^{2}=2-2(-1 / 2)=2+1=3 \Rightarrow n_{d}=\sqrt{3}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355541
If \(\vec{a}+\vec{b}=\vec{c}\) and \(a+b=c\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
1 \(90^{\circ}\)
2 \(180^{\circ}\)
3 \(120^{\circ}\)
4 Zero
Explanation:
Here, \(\vec{a}+\vec{b}=\vec{c}\) and \(c=a+b\) Let \(\theta\) be angle between \(\vec{a}\) and \(\vec{b}\). \(\begin{aligned}& \therefore c=\sqrt{a^{2}+b^{2}+2 a b \cos \theta} \\& a+b=\sqrt{a^{2}+b^{2}+2 a b \cos \theta}\end{aligned}\) Squaring both sides, we get \(\begin{aligned}& (a+b)^{2}=a^{2}+b^{2}+2 a b \cos \theta \\& a^{2}+b^{2}+2 a b=a^{2}+b^{2}+2 a b \cos \theta \\& 2 a b-2 a b \cos \theta=0 \text { or } 1-\cos \theta=0 \\& \cos \theta=1 \text { or } \theta=\cos ^{-1}(1)=0^{0} .\end{aligned}\)
355538
Assertion : The scalar product of two vectors can be zero. Reason : If two vectors are perpendicular to each other, their scalar product will be zero.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(W=F S \cos \theta\), the scalar product of two vectors can indeed be zero. This happens when two vectors are perpendicular to each other as \(\cos 90^{\circ}=0\), in which case their scalar product will be zero. So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355539
If \(\vec{A}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{B}=-\hat{i}+3 \hat{j}+4 \hat{k}\) then projection of \(\vec{A}\) on \(\vec{B}\) will be
1 \(\dfrac{3}{\sqrt{13}}\)
2 \(\dfrac{3}{\sqrt{26}}\)
3 \(\sqrt{\dfrac{3}{26}}\)
4 \(\sqrt{\dfrac{3}{13}}\)
Explanation:
The projection of \(\vec{A}\) on \(\vec{B}=\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}\) \(\begin{gathered}|\vec{B}|=\sqrt{(-1)^{2}+3^{2}+4^{2}}=\sqrt{1+9+16}=\sqrt{26} \\\vec{A} \cdot \vec{B}=2(-1)+3 \times 3+(-1)(4)=3 \\\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}=\dfrac{3}{\sqrt{26}}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355540
If the sum of two unit vectors is a unit vector, then magnitude of difference is
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(1 / \sqrt{2}\)
4 \(\sqrt{5}\)
Explanation:
Let \(\hat{n}_{1}\) and \(\hat{n}_{2}\) are the two unit vectors, then the sum is \(\vec{n}_{s}=\hat{n}_{1}+\hat{n}_{2}\) or \(n_{s}^{2}=n_{1}^{2}+n_{2}^{2}+2 n_{1} n_{2} \cos \theta\) \(=1+1+2 \cos \theta\) Since it is given that \(n_{s}\) is also a unit vector, therefore \(1=1+1+2 \cos \theta \Rightarrow \cos \theta=-\dfrac{1}{2} \therefore \theta=120^{\circ}\) Now the difference vector is \(\hat{n}_{d}=\hat{n}_{1}-\hat{n}_{2}\) or \(\begin{aligned}& n_{d}^{2}=n_{1}^{2}+n_{2}^{2}-2 n_{1} n_{2} \cos \theta=1+1-2 \cos \left(120^{\circ}\right) \\& \quad \therefore n_{d}^{2}=2-2(-1 / 2)=2+1=3 \Rightarrow n_{d}=\sqrt{3}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355541
If \(\vec{a}+\vec{b}=\vec{c}\) and \(a+b=c\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
1 \(90^{\circ}\)
2 \(180^{\circ}\)
3 \(120^{\circ}\)
4 Zero
Explanation:
Here, \(\vec{a}+\vec{b}=\vec{c}\) and \(c=a+b\) Let \(\theta\) be angle between \(\vec{a}\) and \(\vec{b}\). \(\begin{aligned}& \therefore c=\sqrt{a^{2}+b^{2}+2 a b \cos \theta} \\& a+b=\sqrt{a^{2}+b^{2}+2 a b \cos \theta}\end{aligned}\) Squaring both sides, we get \(\begin{aligned}& (a+b)^{2}=a^{2}+b^{2}+2 a b \cos \theta \\& a^{2}+b^{2}+2 a b=a^{2}+b^{2}+2 a b \cos \theta \\& 2 a b-2 a b \cos \theta=0 \text { or } 1-\cos \theta=0 \\& \cos \theta=1 \text { or } \theta=\cos ^{-1}(1)=0^{0} .\end{aligned}\)
355538
Assertion : The scalar product of two vectors can be zero. Reason : If two vectors are perpendicular to each other, their scalar product will be zero.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(W=F S \cos \theta\), the scalar product of two vectors can indeed be zero. This happens when two vectors are perpendicular to each other as \(\cos 90^{\circ}=0\), in which case their scalar product will be zero. So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355539
If \(\vec{A}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{B}=-\hat{i}+3 \hat{j}+4 \hat{k}\) then projection of \(\vec{A}\) on \(\vec{B}\) will be
1 \(\dfrac{3}{\sqrt{13}}\)
2 \(\dfrac{3}{\sqrt{26}}\)
3 \(\sqrt{\dfrac{3}{26}}\)
4 \(\sqrt{\dfrac{3}{13}}\)
Explanation:
The projection of \(\vec{A}\) on \(\vec{B}=\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}\) \(\begin{gathered}|\vec{B}|=\sqrt{(-1)^{2}+3^{2}+4^{2}}=\sqrt{1+9+16}=\sqrt{26} \\\vec{A} \cdot \vec{B}=2(-1)+3 \times 3+(-1)(4)=3 \\\dfrac{\vec{A} \cdot \vec{B}}{|\vec{B}|}=\dfrac{3}{\sqrt{26}}\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355540
If the sum of two unit vectors is a unit vector, then magnitude of difference is
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(1 / \sqrt{2}\)
4 \(\sqrt{5}\)
Explanation:
Let \(\hat{n}_{1}\) and \(\hat{n}_{2}\) are the two unit vectors, then the sum is \(\vec{n}_{s}=\hat{n}_{1}+\hat{n}_{2}\) or \(n_{s}^{2}=n_{1}^{2}+n_{2}^{2}+2 n_{1} n_{2} \cos \theta\) \(=1+1+2 \cos \theta\) Since it is given that \(n_{s}\) is also a unit vector, therefore \(1=1+1+2 \cos \theta \Rightarrow \cos \theta=-\dfrac{1}{2} \therefore \theta=120^{\circ}\) Now the difference vector is \(\hat{n}_{d}=\hat{n}_{1}-\hat{n}_{2}\) or \(\begin{aligned}& n_{d}^{2}=n_{1}^{2}+n_{2}^{2}-2 n_{1} n_{2} \cos \theta=1+1-2 \cos \left(120^{\circ}\right) \\& \quad \therefore n_{d}^{2}=2-2(-1 / 2)=2+1=3 \Rightarrow n_{d}=\sqrt{3}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355541
If \(\vec{a}+\vec{b}=\vec{c}\) and \(a+b=c\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
1 \(90^{\circ}\)
2 \(180^{\circ}\)
3 \(120^{\circ}\)
4 Zero
Explanation:
Here, \(\vec{a}+\vec{b}=\vec{c}\) and \(c=a+b\) Let \(\theta\) be angle between \(\vec{a}\) and \(\vec{b}\). \(\begin{aligned}& \therefore c=\sqrt{a^{2}+b^{2}+2 a b \cos \theta} \\& a+b=\sqrt{a^{2}+b^{2}+2 a b \cos \theta}\end{aligned}\) Squaring both sides, we get \(\begin{aligned}& (a+b)^{2}=a^{2}+b^{2}+2 a b \cos \theta \\& a^{2}+b^{2}+2 a b=a^{2}+b^{2}+2 a b \cos \theta \\& 2 a b-2 a b \cos \theta=0 \text { or } 1-\cos \theta=0 \\& \cos \theta=1 \text { or } \theta=\cos ^{-1}(1)=0^{0} .\end{aligned}\)
